EXAMENSARBETEN I MATEMATIK

EXAMENSARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET The udeiable charm of iequalities Proofs, solutios ad additioal problems av Patri Friggebo 004 - No MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET, 06 9 STOCKHOLM

The udeiable charm of iequalities Proofs, solutios ad additioal problems Patri Friggebo Examesarbete i matemati 0 poäg Hadledare: Paul Vaderlid 004

T ill Jey Kase var vi f ör olia

Cotets Itroductio... 7 Proofs... 8 AM-GM-HM iequality... 8 Chebyshev s iequality... 0 Rearragemet iequality... Cauchy-Schwarz iequality... Hölder s iequality... Miowsi s iequality... 4 Jese s iequality... 5 Power Mea iequality... 7 Schur s iequality... 7 MacLauri s iequality... 8 Muirhead s iequality... 0 Solutios... Additioal problems... 40 Refereces... 4

Itroductio Problems ivolvig the solvig of algebraic iequalities are a frequet category of problems at mathematics competitios. This documet, together with Paul Vaderlid s documet T he U deiable Charm of Iequalities - A Short Guided T our through the Jugle of Algebraic Iequalities, has bee developed to aid studets plaig to participate i such competitios. Here it must be emphasized that this documet is the compaio of Dr. Vaderlid s documet. The algebraic iequalities ad the problems (those with solutios are preseted i his documet, while this documet gives the proofs of the iequalitites ad the solutios to the problems. Hece, i order to optimize this documet, oe must have access to Dr. Vaderlid s documet. As has bee metioed earlier (i Dr. Vaderlid s documet, the iequalities i this documet are ot idepedet ; some ca be derived from others. Readig this documet, oe ca discover some, but ot all, of the relatioships betwee these iequalities. From these relatioships, ad of course also from the more or less edless possibilities usig mathematics, it follow that the solutios to the problems are ot uique. Some problems have several solutios, ad they ca all be foud usig some imagiatio. 7

Proofs. AM-GM-HM iequality Sice the AM-GM iequality is probably the best-ow iequality, ad very useful as well, we will give three proofs of it. Proof (iductio: We start by provig the iequality for =. We have ( a i a 0. We develop the left-had side of this iequality, move the egative term over to the right-had side ad the divide both sides by to get a + a a a. Thus the AM-GM iequality holds for =. Now assume that the iequality holds for some. The we will show that it holds for. ( We have a i = a i + a i a i + a i a i i=+ i=+ a i = a i. i=+ The first iequality i the expressio above holds accordig to our assumptio. The secod iequality follows from the AM-GM iequality for =. Thus by the iductio argumet the AM-GM iequality holds for =, ( =,,... At last we assume that the iequality holds for some. The we will show that it holds for. Let a, a,..., a be positive real umbers ad let a = g = a i. The 8

we have a i + g a i g = g g = g. Hece we have a i + a i a i, i.e. a i ( a i. The iequality i the expressio above holds accordig to our assumptio. Thus the AM-GM iequality holds for all. Proof (by easy aalysis: First we prove oe useful theorem. Theorem : Let a, a,..., a be positive real umbers. Let the α, α,..., α be positive real umbers such that α + α +... + α =. We set G = ad A = α a. The we have G A with equality if ad oly if = a = a =... = a. Proof: We set a = ( + x A. We ote that x >, ( =,,..., ad that α x = 0. Hece G = = = a α = (( + x A α = A = Oe realizes that the iequality A ( + x α A = (+x α A = = e x α = A. = a α e x α holds by studyig the fuctio f(x = e x ( + x. (We have f(0 = 0, ad sice f (x = e x, we have f (x = 0 for x = 0. Fially we have f (0 =, which meas that f(x has miimum for x = 0. Thus, f(x 0 for all x R. We get equality oly whe x = 0, ( =,,...,, i.e. whe a = a =... = a. = Now, if we i Theorem set α =, ( =,,...,, we get 9

a a... + a a + a +... + a, which is the AM-GM iequality. Proof : The AM-GM iequality also follows easily from Jese s iequality. Let a, a,..., a be positive real umbers. We ote that the fuctio f(x = e x is strictly covex, ad that =. Jese s iequality (with x i = l a i, i =,,..., the yields e ( l a + l a +...+ l a el a + el a +... + el a, which after simplifyig becomes the AM-GM iequality. It may also be worth poitig out (as Dr. Vaderlid has already that the AM- GM iequality follows directly from the Power Mea iequality. (Just eep the first ad the last part of that iequality ad you get the AM-GM iequality. Proof of the GM-HM iequality: The GM-HM iequality follows easily from the AM-GM iequality. Let a, a,..., a be positive real umbers. The a, a,..., a are also positive real umbers. Accordig to the AM-GM iequality, we have... a + a +... + a. Whe we ivert both sides of the iequality, the, of course, the iequality gets reversed. Hece a a a a a... a a + a +... +. a. Chebyshev s iequality = We ote that a i b i a i (a i b i a i b j = j= (a j b j a j b i = j= b j = j= (a i b i a i a i b i ( ( a j b j j= j= a i b j = j= b i, ad that a j b i = a j b j j= j= a j b i = 0

= a i b i a i b i. = = It follows that j= a i b i a i b i = (a i b i a i b j + (a j b j a j b i = j= (a i b i a i b j + a j b j a j b i = ((a i a j (b i b j. j= Now we realize that (a i a j (b i b j 0 for i, j =,,...,. (The coditios a i a... a, b i b... b lead to the fact that the two factors o the left-had side of the last metioed iequality above will always have the same sig. Hece, a i b i b i 0. Movig the right term o the a i left-had side of this iequality to the right-had side ad the dividig both sides by yields Chebyshev s iequality. It is ot hard to realize that i Chebyshev s iequality we will have equality if ad oly if a = a =... = a or b = b =... = b This follows from the iequalitites (a i a j (b i b j 0, i, j =,,...,. To realize that the iequality is reversed if we have a i a... a ad b i b... b is o more difficult. This follows from the iequalities (a i a j (b i b j 0, i, j =,,..., (these iequalities are reversed because ow the coditios lead to the fact that the two factors o the left-had side will always have the opposite sig. Aother proof: We will ow show that Chebyshev s iequality easily ca be derived from the Rearragemet iequality. Chebyshev s iequality ca be expressed as (a + a +... + a (b + b +... + b (a b + a b +.. + a b. O the left-had side of this iequality we multiply the two parethesis together. We the get terms o that side. The we arrage these terms i groups as follows: (a b +a b +...+a b +(a b +a b +...+a b +...+(a b +a b +...+a b. Accordig to the Rearragemet iequality a b + a b +.. + a b ayoe of j=

these groups of umbers, ad Chebyshev s iequality follows. Usig the same techique oe ca easily show the case whe Chebyshev s iequality is reversed.. Rearragemet iequality We start by showig the iequality for =. Let a a ad b i b be real umbers. The we have (a a (b b 0, sice both factors o the left-had side of the iequality are defiitely oegative. Multiplyig these factors together ad the rearragig the terms yields the iequality a b + a b a b + a b, which is the Rearragemet iequality for =. We realize that we have equality oly if a = a or b = b. Now we prove the iequality i the geeral case. Let b b... b ad c, c,..., c be real umbers. Let the a, a,..., a be a permutatio of c, c,..., c such that a b +a b +...+a b is maximal. Now, assume that we for some i < j have a i > a j. The we have a i b j + a j b i a i b i + a j b j (the case =. Thus, a b + a b +... + a b is ot maximal uless a a... a or b i = b j for all i < j such that a i > a j. I the later case the umbers a i, a j ca chage places so that we get a a... a. Hece, a b + a b +... + a b is maximal whe a a... a. Now we ote that (a b +a b +...+a b = ( a b +( a b +...+( a b is maximal whe a a... a ad b b... b. From that follows that a b + a b +... + a b is miimal uder the same coditios. But the coditio a a... a is equivalet to the coditio a a... a. Thus, a b + a b +... + a b is miimal whe b b... b ad a a... a. 4. Cauchy-Schwarz iequality Let a, a,..., a ad b, b,..., b be real umbers. Let us start with the polyomial (a i x + b i = x + a i b i ( ( x + b i. a i Sice (a i x + b i 0 for all real umbers x, it is obvious that

( a i ( x + a i b i x + b i 0 too. It follows that the discrimiat for a i this polyomial is 0. The discrimiat for a quadratic polyomial ax + bx + c ( is b 4ac. Thus, 4 a i b i 4 b i 0. If we the move the righthad term o the left-had side of this iequality to the right-had side ad fially divide both sides by 4, we ed up with the Cauchy-Schwarz iequality. So, why does the discrimiat for the above polyomial have to be 0? If we solve the equatio ax + bx + c = 0 for arbitrary real umbers a, b, c usig quadratic completio, we get the solutio x = b a ± b 4ac. Now, if we a have ax + bx + c 0, we ow that the above equatio will have o real roots (if we have ax + bx + c > 0 or oe double root (if we have ax + bx + c = 0. If we have o real roots, the sum of the terms uder the root sig will be egative, i.e. b 4ac < 0. If we have a double root, we will have b 4ac = 0. It follows from the fact that (a i x + b i 0 for all real umbers x, that the Cauchy-Schwarz iequality will have equality if ad oly if the umber sequeces a ad b are proportioal. 5. Hölder s iequality Set p = s ad q =. The Hölder s iequality becomes t ( ( s t a b. The iequality ca the be expressed the = followig way: = = a s ( = = a s as i s b t ( = b t bt i t. We ote that s + t =. The, applyig Theorem (see proof of the AM-GM iequality to the left-had side of this iequality yields ( a s ( s b t ( t a s as i bt i s + b t as i t = bt i s + t =. We have equality if ad oly if the umber sequeces (a s, a s,..., a s ad (b t, bt,..., bt are proportioal. This follows from the coditios for equality i

Theorem. We will ow show a additioal useful iequality: If we i Hölder s iequality allow oe of the parameters p, q to be egative, the the iequality is reversed. Suppose that p < 0, i.e. s < 0 with the otatio above. We set S = s t ad T = t. The we have S > 0, T > 0 ad S + T =. Now we set u = a t v = a t bt, ( =,,...,. The Hölder s iequality yields ( u v = = a t at bt = u S S ( = ( = a t ( s t v T t s T, which of course is equivalet to ad ( t. (a t bt t After some simplifyig ad = rearragig ad the taig the t:th root of both sides of the iequality, we get ( ( s t a b, which is the reversed iequality. = = = a s = 6. Miowsi s iequality We start by otig that a i (a i + b i r + r r b t (a i + b i r = (a i + b i (a i + b i r = b i (a i + b i r. For r > we set s so that r + =, i.e. s s =. The, by Hölder s iequality, we have the followig (we apply the iequality to both of the terms a i (a i + b i r ad b i (a i + b i r : (a i + b i r ( = ( a r i a r i r r ( (a i + b i (r s ( s + ( (a i + b i r r r divide both sides of this iequality by ( + b r i b r i r r ( (a i + b i (r s s = ( (a i + b i r r r. If we the ( r (a i + b i r r, we get Miowsi s 4

iequality. It is obvious that Miowsi s iequality holds for r = ; that coditio yields equality. Whe r > we have equality if ad oly if the umber sequeces a ad b are proportioal. This follows from the coditios for equality i Hölder s iequality. If r <, r 0, the s becomes egative, which leads to Miowsi s iequality gettig reversed. This follows from the fact that Hölder s iequality becomes reversed for s < 0. 7. Jese s iequality We will prove Jese s iequality for ratioal positive umbers α, α,..., α oly. The geeral proof for oegative real umbers α, α,..., α requires some serious aalytical argumets. We will divide the proof of Jese s iequality ito two cases. Case : α i =, (i =,,...,. (We use the same approach as i the first proof of the AM-GM iequality. I this case we are supposed to prove the iequality ( ( f x i f(x i. The iequality is true for =. This follows directly from the defiitio of covexity. Now, assume that ( holds for =, ( =,,... The ( also holds for m = + =. Proof: Let x, x,..., x m I. The we have ( x + x +... + x ( m f = f x i + x +i ( m ( f x i + f x +i f(x i + f(x +i = m = f(x i. m I the expressio above, the first iequality holds because ( holds for 5

=. (We have that the two umbers x i, x +i I, sice they are the arithmetic averages of umbers that all I. The secod iequality follows from our assumptio. Thus, sice ( holds for =, by iductio it holds for =, ( =,,... Assume that ( holds for >. The ( also holds for Proof: Let x, x,..., x I. For these umbers together with the umber x = (x + x +... + x (the arithmetic average of x, x,..., x, ( holds accordig to our assumptio. We get ( x + x +... + x + x +x +...+x ( f f(x + f(x +... + f(x + f( x +x +...+x. ( x + x +... + x After simplifyig, the left-had side of ( becomes f. Thus, ( x + x +... + x f f(x i + ( f x + x +... + x. Further ( x + x +... + x simplifyig yields f f(x i. Now, by iductio, ( holds for. Hece, ( is prove i case. Case : α, α,..., α are ratioal positive umbers. Sice α, α,..., α are ratioal positive umbers, there is a atural umber m ad oegative itegers p, p,..., p such that m = p +p +...+p ad α i = p i m, (i =,,...,. (To realize this, just rewrite the umbers α, α,..., α so that they all have the same deomiator. Case yields ( (x +... + x +... + (x +... + x ( f m (f(x +... + f(x +... + (f(x +... + f(x. (The first parethesis i m 6

the omiator o the left-had side of the iequality cotais p terms, the secod parethesis p terms, ad so o. Now, ( ca be expressed as ( f p i x i p i f(x i. Hece, ( is prove i case. m m 8. Power Mea iequality We start by poitig out the obvious: for = m we have equality, which meas that the iequality holds i that case. Now we assume istead that < m. The we have m > ad it follows that the fuctio f(x = x m is strictly covex for x 0. (The secod derivative, f (x = m ( m x m > 0 for x > 0. Now, sice the umbers a, a,..., a are oegative, clearly the umbers a, a,..., a are oegative too. With help from Jese s iequality we the get the iequality ( (a m + (a m +... + (a m a + a +... + m a, which, after some simplifyig, becomes ( m (am + am +... + am (a + a +... + a. Taig the m:th root of both sides of this iequality the yiel ds the Power Mea iequality. The coditio for equality follows from the coditio for equality i Jese s iequality. 9. Schur s iequality We will prove Schur s iequality by provig a stroger theorem. Theorem : If a, b, c, u, v, w are oegative real umbers ad we have ( a p + c p b p ad ( u p+ + w p+ v p+ the, if p > 0, we have ( ubc vca + wab 0. Proof: We start with two pairs of oegative umbers: (a p+, c ((uc p+, (wa p+. Sice we have p > 0, we have p + > 0, p p + + p + (a p+ p+ p + c p+ ad p p + > 0 ad =. The Hölder s iequality yields a p+ (uc p+ + c p+ (wa p+ p+ p+ p p p+ ((uc p+ (p+ + (wa p+ (p+ p+, i.e 7

(ac p p+ u p+ + (ac p+ w p+ (a p + c p p+ (uc + wa p+. Taig the (p + :th power of both sides of this iequality, we get ac(u p+ + w p+ p+ (a p + c p p (uc + wa. Now we use the coditios ( ad ( i order to get ac(v p+ p+ (b p p (uc + wa, which is equivalet to ( ubc vca + wab 0. Now we ca assume that 0 z y x. The, usig Theorem (settig p =, a = y z, b = x z, c = x y, u = x r, v = y r, w = z r, we get x r (x z(x y y r (x y(y z + z r (y z(x z 0, which is equivalet to Schur s iequality. It is quite easy to show that we have equality if ad oly if x = y = z. The oly oe of the three terms o the left-had side of Schur s iequality that ca be egative is y r (y x(y z. It is egative whe we have y < x ad y > z. But with those coditios we see that x r (x z(x y > y r (y x(y z. Thus, to have equality we must have y = x or y = z. I either of these two cases two of the three terms o the left-had side of the iequality = 0. It follows that we must have x = y = z to have equality. 0. MacLauri s iequality We start by provig a theorem that we eed i order to prove MacLauri s iequality. Theorem : For, let a, a,..., a be positive real umbers that are ot all a i a i...a ir equal. Also let p 0 = ad p r = (, r =,,...,. The, r r, r <, p r p r+ < p r. i<i<...<ir Proof (iductio over : For = we have p 0 =, p = a + a ad ( a + a p = a a. Thus, p 0 p = a a < = p. Of course the iequality above follows from the AM-GM iequality (sice we have a a, we have a strict iequality. Hece we have prove the theorem for =. Suppose ow that the theorem holds for some =,. We will show that it the holds for =. 8

Accordig to our assumptio, for the positive real umbers a, a,..., a that are ot all equal we have P r P r+ < Pr, r <. (Here we write P istead of p to be able to separate these umbers that form the iequality for = from the umbers that form the iequality for = ; the iequality that we are supposed to prove. We wat to show that for the positive real umbers a, a,..., a that are ot all equal, the iequality p r p r+ < p r, r <, holds. Now observe that p r = r P r + r a P r, r =,,...,, where we set P( = 0. ((This ca be ( a little tricy to realize, but it follows from the fact that = +. We get ( (p r p r+ p r = = (( r + P r + r ( r a P r P r+ + r + a P r ( r P r + r a P r = A + Ba + Ca, where A = (( r P r P r+ ( r P r, B = ( r +(r +P r P r +( r (r P r P r+ ( rrp r P r, C = (r P r P r r P r. Sice ot all a, a,..., a are equal, we have P r P r+ < Pr, P r P r < Pr ad P r P r+ < P r P r. (The last iequality holds because P r+ P r P r+ = P r P r < P P r+ r = P r P r P r+ < P r Pr = P r P r. P r P r P r P r From these iequalities follow that A < P r, B < P r P r ad C < P r. Furthermore, from ( we get (p r p r+ p r < P r + P rp r a P r a = (P r P r a 0. Thus, (p r p r+ p r < 0, which is equivalet to p r p r+ < p r. Hece the theorem is prove for all. Oe more thig remais, though; the case where 9

a = a =... = a a. I that case we have a = P r, ad from ( we get P r (p r p r+ p r = a P r + a Pr a Pr a = (a a Pr < 0. Thus the theorem is prove i that case also. Proof of MacLauri s iequality: Cosider the iequalitites p 0 p < p (p p < p 4 (p p 4 < p 6... (p r p r+ r < p r r If we multiply together all the left-had ad right-had sides respectively, we come up with the iequality p 0 p p 4 p 6...pr r pr r p r r+ < p p 4 p 6...pr r pr r, which yields p r r+ < pr+ r. The last iequality ca also be writte as r pr > r+ pr+. Hece, we have p > p > p >... > p > p, which is MacLauri s iequality. I the case where a = a =... = a, we have p = p = p =... = p = p. This is easy to show. I that case we ( ( have p = ( a = a = ( a = ( = ( a = p.. Muirhead s iequality We start by provig the followig lemma: Lemma: Let a, a, b, b be oegative real umbers such that a + a = b + b ad max(a, a max(b, b. Let also x, y be oegative real umbers. The we have 0

( x a ya + xa ya xb yb + xb yb. Proof: Because of symmetry we ca without loss of geerality assume that a a ad b b. If ay of the umbers x, y = 0 the ( obviously holds. Therefore we assume that x, y 0. We have x a ya + xa ya xb yb xb yb = = x a y a (x a a + y a a x b a y b a x b a y b a = = x a ya (xb a y b a (x b a y b a 0. Now, why does x a y a (x b a y b a (x b a y b a 0 hold? Clearly, we have x a ya 0. The, from the coditios the iequalities b a 0 ad b a 0 follow. If we the have x y, we have x b a y b a 0 ad x b a y b a 0, ad the iequality follows. If we o the cotrary have x y, we have x b a y b a 0 ad x b a y b a 0, ad the iequality follows i that case too. We divide the proof of Muirhead s iequality ito two cases. Case : b a. Accordig to the coditios, we have a a + a b ad a b. Thus, max(a, a max(a +a b, b. We also ote that a +a = (a +a b +b. We use the lemma ad get x a y a + x a y a x a +a b y b + x b y a +a b. Clearly also z a (xa ya +xa ya za (xa +a b y b +xb ya +a b holds. The we realize that ( z a (x a y a + x a y a z a (x a +a b y b + x b y a +a b holds too. cyclic cyclic (Whe we used the lemma, we could for example have used the umbers y ad z istead of the umbers x ad y ad the multiplied both sides of the iequality with x a istead of za. Now, we ote that cyclic z a (x a y a + x a y a = sym x a y a z a. Accordig to the coditios, we also have a + a b b b, ad max(a + a b, a max(b, b follows. We ote that (a + a b + a = b + b. Usig the lemma yet aother time yields y a +a b z a + y a z a +a b y b z b + y b z b. Aalogous to previous argumets oe the realizes that

( x b (y a +a b z a + y a z a +a b x b (y b z b + y b z b also holds. cyclic cyclic The we ote that x b (y b z b + y b z b = x b y b z b. At last we ote that cyclic sym the right-had side of ( is equal to the left-had side of (. The, from ( ad (, Muirhead s iequality follows. Case : b a. Accordig to the coditios, we have b b + b + b = a + a + a b + a + a. From this follows that b a + a b. Accordig to the coditios, we also have a b. Hece, max(a, a max(b, a + a b. We ote that a + a = b + (a + a b. Now, we use the lemma to get y a za + ya za y b za +a b + y a +a b z b. Aalogous with case, we see without problem that ( x a (y a z a + y a z a x a (y b z a +a b + y a +a b z b holds. We cyclic cyclic ote that cyclic x a (y a z a + y a z a = sym x a y a x a. Accordig to the coditios, we also have max(a, a + a b max(b, b. We ote that a + (a + a b = b + b ad use the lemma oe fial time to get x a za +a b + x a +a b z a xb zb + xb zb. The we ote that (4 y b (x a z a +a b + x a +a b z a y b (x b z b + x b z b holds. We cyclic cyclic see that y b (x b z b + x b z b = x b y b z b. Aalogous with case we see cyclic sym that the right-had side of ( is equal to the left-had side of (4. Thus, from ( ad (4, Muirhead s iequality follows.

Solutios. Usig the hit, we realize that (x + y (y + z (z + x (x + 4y(y + 4z(z + 4x = = (x + y + y + y + y(y + z + z + z + z(z + x + x + x + x. The, applyig the AM-GM iequality to each of the parethesis, we get (x + y + y + y + y(y + z + z + z + z(z + x + x + x + x 5 5 xy 4 5 5 yz4 5 5 zx4 = 5xyz.. Applyig the AM-GM iequality to the deomiator of each term o the left-had side of the iequality, we get x x 4 + x + y y 4 + x x x 4 y + y y 4 x = xy + xy = xy.. We apply the AM-GM iequality to the left-had side of the iequality ad get a + 4 b + 4 c + 6 56 d 4 4 abcd = 4 6. The, usig the AM-GM iequality abcd 6 agai, this time o the deomiator, we get 4 6 64 = abcd a+b+c+d a + b + c + d. 4 ( 4. We have + a ( + b ( + c = b c a = + c a + b c + b a + a b + c + a c + = ( a = + a + a b + a ( b + c b + b a + b ( c + c c + c a + c. The, applyig the AM-GM b iequality to each of the parethesis, we get ( a + a + a b + a ( b + c b + b a + b ( c + c c + c a b + c we have to show is that + a + b + c ( abc + a + b + c. Now, all abc + a + b + c abc, which is the same as showig that a + b + c. This iequality is obviously true; we just apply the abc AM-GM iequality to the deomiator to get a + b + c a + b + c abc =. (a + b + c 5. We start, as suggested, by showig that x + y + z xy + yz + zx. We

mae the variable substitutio x = a, y = b ad z = c. The we have to show that a + b + c ab + bc + ca. Usig the AM-GM iequality o each of the terms o the right-had side of the iequality, we get a + b ab + bc + ca + b + c + c + a = a + b + c. (Those familiar with the Rearragemet iequality may thi that the proof above is superfluous, sice the prove iequality is a simple cosequece of the Rearragemet iequality. Now we ca show that x +y +z +xy+yz+xz ( x+ y+ z. By usig our ( hit iequality, we see that x + y + z + xy + yz + xz (xy + yz + xz = xy + xz xy + yz xz + yz = + +. The we apply the AM-GM iequality to ( xy + xz xy + yz xz + yz each term iside the parethesis to get + + ( x xyz + y xyz + z xyz = ( x + y + z = = ( x + y + z ( x + y + z. 6. We start with the hit y i = i y. Usig the AM-GM iequality o the right-had side of this iequality, we get y i ( i y. Of course this is true for all i (i =,,...,, which meas that we have iequalities lie the oe above. We multiply together all the right-had sides of these equalities ad treat similarly all the left-had sides. The we come up with the iequality ( y i ( y i. Dividig both sides by y i yields the iequality y i y i (. Now, y i = 998 x i +998 y 998 = x i + 998 998 i x i 998 +998 Thus, our iequality becomes = x i 998. rearragig becomes the iequality that we set out to prove. 7. For proof of the hit iequality, see problem 5. x i 998 (, which after some easy Usig the AM-HM iequality o the left-had side of the iequality, we get 4

+ ab + + bc + + ca 9. The, applyig the hit iequality, + ab + bc + ca 9 we get + ab + bc + ca 9 + a + b + c = 9 6 =. 8. We start by showig the iequality t + t (t > 0. We apply the AM-GM iequality to the left-had side of this iequality to get t + t t t =. Now to the mai iequality. We ote that all the deomiators are positive. The we subtract x from both sides of the iequality ad express the left-had side of the iequality i a clever way. We ow have to prove the iequality (x 0 x + + (x x + +... + (x x +. x 0 x x x x x This becomes easy with the help of our hit iequality. The left-had side of the above iequality cosists of terms. If we add the two first terms together, ad the add the two followig terms together, ad so o, the, accordig to our hit iequality, the above iequality holds. 9. We start by multiplyig both sides of the equality by (4x + y. The we calculate the right-had side of the equality ad come up with the ivitig equality 64x y = 8x 4 y + x y + 4x y + xy 4 + 6x 4 + 4x y + 8x y + xy + 8x y + xy + 4x y + y 4 + 6x + 4xy + 8x y + y. This is, of course, the same coditio with which we bega. The, usig the AM-GM iequality o the right-had side of the equality, we get 8x 4 y + x y + 4x y + xy 4 + 6x 4 + 4x y + 8x y + xy + 8x y + xy + 4x y + y 4 + 6x + 4xy + 8x y + y 6 6 4 4 x y = 64x y. What a coicidece! Now, the above iequality becomes a equality if ad oly if all the terms o the left-had side are equal. To fid the pairs of positive real umbers x, y that we wat, we simply choose two suitable terms, set them equal to each other, ad the solve the equatio. The equatio 8x 4 y = 8x y gives x =. The equatio 8x y = 6x gives y =. Checig all the terms, we realize that they are all equal for x =, y =. 5

Thus, we see that the oly pair of positive real umbers for which the coditio above holds is x =, y =. 0. We follow the hit ad apply the AM-GM iequality three times; to the left-had side of the equality o the factors ( + ax = ( + ax + ax ad ( + by = ( + by + by, ad the to the whole right-had side. We get 7xy ( + ax( + by 7xy xy a x b y = a b ad ab + x a + y xy b a b. Now, to have equality i the two iequalities above, the coditio for equality i the AM-GM iequality has to be met. This coditio leads to ( ax = by = ad ( ab = x a = y b. ( yields x = a, y = b while from (, we get x = b, y = a. Hece, if we have a b, the there is o pair of positive real umbers x, y such that 7xy ( + ax( + by = ab + x a + y. But if we have a = b, the we have equality b for x = y = a.. We start by multiplyig both sides of the iequality by (a + b + c. The we express that iequality i a clever way. We arrive at the iequality a a + b b + c c a a + b b + c c + a a + b b + c c + a b + b c + c a + a a + b b + c c + a c + b a + c b. Becau- se of symmetry, we ca, without loss of geerality, assume that a b c. It follows that a b c. The, usig the Rearragemet iequality, we realize that ay term o the left-had side of the iequality (OK, they are idetical is ay term o the right-had side of the iequality. Thus the above iequality holds.. We log both sides ad the use the well-ow log laws to rewrite the iequality. We get a l a + b l b + c l c a + b + c l a + a + b + c l b + a + b + c l c. Because of symmetry, we ca oce agai without loss of geerality assume that 6

a b c. It follows that l a l b l c. Now we ca apply Chebyshev s iequality to the left-had side of the iequality. We get a l a + b l b + c l c a + b + c (a+b+c(l a+l b+l c = l a+ a + b + c l b+ a + b + c l c.. The left-had side of the iequality ca be rewritte as a +a +...+a. It is obvious that.... The, accordig to the Rearragemet iequality, a + a +... + a is miimal whe we have a < a <... < a ; that is whe we have a i = i, (i =,,...,. Thus we have a + a +... + a + +... + = + + +. 4. The left-had side of the iequality ca be rewritte as a +a +...+a. To miimize this expressio with help from the Rearragemet iequality, the largest of the umbers a, a,..., a has to be a a a multiplied by the smallest of the umbers a, a,..., a, the secod largest multiplied by the secod smallest, ad so o. Now, suppose that of all the umbers a, a,..., a, the umber a is the largest. Which oe of the umbers a, a,..., a is the the smallest? a of course! If the umber a l is the secod largest the a l is the secod smallest, ad so o. Thus a + a +... + a a + a +... + a =. a a a a a a 5. We have a8 + b 8 + c 8 = a 5 a b c b c +b5 a c +c5 a b. Because of the symmetry, we ca without loss of geerality assume that a b c. From that follows that a 5 b 5 c 5 ad a b a c b c. The, usig the Rearragemet iequality twice, we get a 5 b c + b5 a c + c5 a b a5 a b + b5 b c + c5 a c = = a b + b c + c a a a + b b + c c = a + b + c. 6. We have a x (by + cz(bz + cy + b y (cz + ax(cx + az + c z (ax + by(ay + bx = 7

a x = b y z + by c + cz b + c z y + b y c z x + cz a + ax c + a x z + c z. We tae a loo at the deomiators ad ote a x y + ax b + by a + b y x that, accordig to the coditios, the followig is true: The deomiator of the first term: b y c z, z y, by cz, c b. The deomiator of the secod term: a x c z, z x, ax cz, c a. The deomiator of the third term: a x b y, y x, ax by, b a. We apply the Rearragemet iequality to the deomiators ad get a x b y z + by c + cz b + c z y + b y c z x + cz a + ax c + a x z + c z a x y + ax b + by a + b y x a x b y y + by b + cz c + c z z + b y c z z + cz c + ax a + a x x + c z. We ow a x x + ax a + by b + b y y mae the suggested variable substitutio. We are ow left to prove that the iequality ( α β + γ + β γ + α + γ holds. We start this proof by multiplyig α + β 4 both sides of the iequality by (α(β + γ + β(γ + α + γ(α + β. The iequality that we get from that, we deote by (. The we apply the Cauchy-Schwarz iequality to the left-had side of ( to get ( α β + γ + β γ + α + γ (α(β + γ + β(γ + α + γ(α + β α + β ( α β + γ α(β β + γ + γ + α β(γ γ + α + α + β γ(α + β = = (α+β +γ = α +β +γ +(αβ +αγ +βγ. Now, looig at the right-had side of (, we see that (α(β +γ+β(γ +α+γ(α +β = (αβ +αγ +βγ. 4 Thus to complete our proof, we just eed to prove the iequality α + β + γ αβ + αγ + βγ. For this simple proof, see problem 5. 7. We ote that ((a + b + (b + c + (c + d + (d + a = ad multiply both sides of the iequality by this factor to get the iequality ( a ((a + b + (b + c + (c + d + (d + a a + b + b b + c + c c + d + d d + a 8

. Usig the Cauchy-Schwarz iequality o the left-had side of this iequality ( a yields ((a + b + (b + c + (c + d + (d + a a + b + b b + c + c c + d + d d + a ( a b a + b a + b + c b + c b + c + d c + d c + d + d + a = d + a = (a + b + c + d =. To realize that we have equality if ad oly if a = b = c = d =, we start by 4 otig that whe we have equality, the umber sequeces ( a + b, b + c, c + d, ( a d + a ad b a + b, c b + c, d c + d, are d + a proportioal. Thus whe we have iequality there is (sice all the umbers i these umber sequeces are positive a umber > 0 such that ( a a + b = a + b, ( b b + c = b + c, ( c c + d = c + d ad (4 d d + a = d + a. These four equatios yield i tur = a, = b, = c ad = d. Hece, whe we have equality, we have a = b = c = d = 4. 8. We start, as suggested, by multiplyig both sides of the iequality by the factor (a(b + c + b(c + a + c(a + b. We deote the iequality that we get from that by (. Usig the Cauchy-Schwarz iequality o the left-had side of (, we get ( a (a(b + c + b(c + a + c(a + b b + c + b c + a + c a + b ( a b c a(b + c b + c + (b(c + a c + a + c(a + b = a + b = (a + b + c = a + b + c + (ab + ac + bc. Worig out the right-had side 9

of (, we get a(b + c + b(c + a + c(a + b = (ab + ac + bc. Oce agai we ed up havig to prove the iequality a + b + c ab + ac + bc. See problem 5 for that proof. 9. We start by multiplyig both sides of the iequality by (ac + bd to get the ( a iequality (ac + bd c + b ac + bd. The, applyig the Cauchy-Schwarz d iequality to the left-had side yields ( a ( (ac + bd c + b ac a d c + b bd = (a + b. We ow have d to show that the iequality (a + b ac + bd holds. We square this iequality. The we get (a + b 4 = (a + b (a + b = (a + b (c + d (ac + bd. The iequality (a + b (c + d (ac + bd holds, of course, accordig to the Cauchy-Schwarz iequality. 0. For this problem we will give two solutios. Solutio (usig Hölder s iequality: We mae the suggested variable substitutio. The iequality to be prove becomes (+a (+a...(+a (+a a...a. We tae the :th root of both sides ad the write the iequality as ( +a ( +a...( +a (+a a...a. It is obvious that this iequality holds (Hölder s iequality for umber sequeces. Solutio (usig the AM-GM iequality: Usig the biomial theorem, we develop both sides of the iequality. We get the iequality + (x ( + x +... + x + ((x x + x x +... + x ( x +... + (x x...x + (x x...x + (x x...x +... + (x x...x. We subtract from both sides of the iequality. The we have parethesis o the left-had side of the iequality, ad terms o the right-had side. If we are able to show that the :th parethesis o the left-had side ( =,,..., is the :th term o the right-had ( side, we are doe. Now, the :th parethesis o the ( left-had side cosists of terms. The factor a, ( =,,..., exists i of ( these terms. We use the AM-GM iequality o these terms to get that the ( ( parethesis is x ( x (...x ( (. We ote that 0

( = (! (!(!, ad that!(! ( = =! ( ( ( Hece, we get x ( x (...x ( ( = (x x...x. Sice ( (x x...x is the :th term o the right-had side of the iequality, we are doe. (!(!. (!. Usig Hölder s iequality (with p = ad q = o the right-had side of the iequality, we get a b + b c + c a ( (a + (b + (c (b + c + a = a + b + c.. Applyig Hölder s iequality for three umber sequeces (these sequeces beig (a p, b p, c p, (a q, b q, c q ad (a r, b r, c r to the right-had side of the iequality yields a p b q c r + c p a q b r + b p c q a r ( (a p p + (b p p + (c p p p = (a + b + c (p+q+r = a + b + c. ( (b q q + (a q q + (c q q q ( (c r r + (b r r + (a r r r =. We start by taig the third root of both sides of the iequality. The we express x ad (( as (x x ad respectively. We come up with the i- x ( equality (x ( = = (( x. Usig first Hölder s iequality o the left-had side of this iequality ad the usig the coditio x [, ], ( =,,...,, we get (x ( (( x = ( ( ( x = ( x b = c = = 4. We set a = y a a + b + c, b = z a a + b + c, c = = x a a + b + c, a = x z b a + b + c, b = x b a + b + c, c = = ( = y b a + b + c, a = x c a + b + c ad y c a + b + c, =. z c a + b + c,

ad use Miowsi s iequality for three umber sequeces. We get ax + by + cz ay + bz + + cx az + bx + + cy a + b + c a + b + c a + b + c (a + b + c + (a + b + c + (a + b + c = (x a + y a + z a ( y b + z b + x b ( z c + x c + y c, = + + a + b + c a + b + c a + b + c which after some simplificatio becomes x + y + z. 5. We start by rearragig the iequality a little. We get the equivalet iequality ( a + b + 4 c + 6 d a + b + c + d 64. We use Hölder s iequality o the left-had side of this iequality to get ( a + b + 4 c + 6 d a + b + c + d ( a a + b + c + 4 d = 8 = 64. b c d 6. First we show that the left iequality holds. We ote that f(x = x is strictly covex for x > 0 (we have f (x = > 0 for x > 0. The we divide x both sides of the iequality by 6 to come up with the iequality, 5 a + b + c a + b + b + c +. Usig Jese s iequality (with c + a the fuctio metioed above o the right-had side of the iequality the yields a + b + b + c + c + a (a + b + (b + c + = (c + a = a + b + c =, 5 a + b + c. To show that the right-had iequality holds we start by expressig the righthad side of this iequality i a clever way. The, oce agai, we use Jese s iequality; this time o each of the parethesis (the fuctio beig the same as before. We get a + b + ( c = a + + b a + b + b + c + ( c + a = ( b + c a + b + b + c + c + a ( +. c + a

7. We start by taig a loo at the left-had side of the iequality. We rearrage x this expressio, gettig x + + y y + + z z + = = x + x + +y + y + +z + = z + x + y +. Our iequality z + to be prove is thus the followig: x + y + z +. After some 4 easy rearragig we come up with the equivalet iequality 4 ( x + + y + +. We ote oce more that f(x = is strictly z + x covex for x > 0. The we apply Jese s iequality to the right-had side of the last iequality. We get ( x + + y + + z + (x + + (y + + (z + = + x + y + z = 4. 8. We set f(x = 4x x. If oe taes a loo o f (x, oe realizes that we have f (x > 0 for 0 < x <, which meas that f(x is strictly covex i this iterval. We ca the use Jese s iequality. We get the followig: 4a a = 4a a 4( = a ( = a = = = = 4( ( =. Thus, the miimum value of the expressio is. 9. If oe draws a arbitrary triagle with a iscribed circle of radius ad the comes up with some trivial symmetry argumets, the oe realizes that the hit holds. Now, the fuctio f(x = ta x is strictly covex for 0 < x < π. This, i tur, oe realizes by studyig f si x (x =. Usig Jese s iequality the yields ta α + ta β + ( ta γ = ta( = (α + β + γ ta π = ta π 6 6 circumferece of the equilateral triagle. si 4 x ta α + + ta π 6 ta β + ta γ + ta π, which is half the 6 0. We mae the suggested variable substitutio to get the iequality e a e e a + 8 + b e e b + 8 + c. The coditios 0 < x, y, z < 4 ad xyz = e c + 8

become the coditios 0 < e a, e b, e c < 4 ad a + b + c = 0 respectively. The we ( e s study the fuctio f(s =. After some wor we get f 4e s (s = e s + 8 (e s + 8 ad f (s = e s (8 e s. We see that f (s > 0 for e s < 4. Hece f(s is (e s + 8 5 strictly covex for e s < 4. Now we multiply both sides of our iequality by. We e a e b e c get e a + 8 + e b + 8 + e c + 8. Applyig Jese s iequality to the left-had side the yields e a e a + 8 + e b e b + 8 + e c e c + 8 e (a+b+c e (a+b+c + 8 = + 8 =, ad we are doe.. We use the hit ad fid, accordig to the Power Mea iequality, that ( x + y ( x + y, which is equivalet to the iequality x + y (x + y. Applyig the coditio x + y > to the right-had side of this iequality yields (x + y < (x + y (x + y = x + y. Thus we have ( x + y < x + y. We also have ( y 0. Developig the left-had side of this iequality ad the multiplyig both sides by y yield ( y y + y 4 0. Now, if we add the left-had ad the right-had sides of ( ad ( respectively, we come up with the desired iequality.. We start by showig that the iequality a + a + a + a 4 a + a + a + a 4 holds. We divide both sides of the iequality by 4. The we apply the Power Mea iequality to the right-had side, gettig 4 (a + a + a + a 4 ( 4 (a + a + a + a 4. What we ow have to show is the iequality 4 (a + a + a + a 4 ( 4 (a + a + a + a 4. 4

It is obvious that this iequality holds if the iequality 4 (a + a + a + a 4 holds. Ad it does; usig the AM-GM iequality o the left-had side yields 4 (a + a + a + a 4 4 a a a a 4 =. The secod iequality that we have to show is a + a + a + a 4 a + a + a + a 4. To show that, we set A = a + a + a + a 4 ad A i 4 4 A i = 4A a i = A. Thus, = A a i, (i =,,, 4. We see that 4 A i = A. Now we ote that, by the AM- GM iequality, we have A = (a +a +a 4 a a a 4 = a a a 4 = a. Usig the same reasoig, we realize that the iequalities A a, A a ad A 4 a 4 also hold. Addig the left-had ad the right-had sides respectively of these four iequalities the yields the desired iequality.. We start by showig the hit iequality. We multiply both sides of the hit iequality by. Applyig the Power Mea iequality to the right-had side of the iequality the yields (a + b + c (a + b + c ( (a + b + c ( ((a + (b + (c = (a + b + c. Now to the iequality that we are supposed to show. Usig the hit iequality o the left-had side of that iequality, we get a + b + c + a + b + d a + b + c a + b + d + a + c + d a + c + d + b + c + d b + c + d (a + b + c(a + b + c + (a + b + d(a + b + d + a + b + c (a + c + d(a + c + d a + c + d + a + b + d (b + c + d(b + c + d b + c + d = (a + b + c + (a + b + d + (a + c + d + (b + c + d = = a + b + c + d. 4. We follow the hit ad homogeize the iequality. O the left-had side of = 5

the iequality we multiply the term xyz by x + y + z. The we multiply the right-had side by (x + y + z. We develop the right-had side ad simplify the iequality to come up with the iequality xyz x + y + z xy+xz+yz. We cotiue by squarig both sides ad the gettig rid of terms that cacel each other. We get the iequality x yz + xy z + xyz x y + x z + y z. To see that this iequality holds, we mae the variable substitutio xy = a, xz = b, yz = c. This leads to the, probably by ow, familiar iequality ab + ac + bc a + b + c, which holds. For proof of that, see problem 5. 5. We start by multiplyig the left-had side of the iequality by (p + q + r. The, o the right-had side of the iequality, we multiply the term by (p + q + r to get a homogeous iequality. We develop all the terms to ed up with a iequality with ridiculously may terms. Fortuately, may of the terms cacel out each other. We ed up with the iequality p q + pq + p r + pr + q r + qr (p + q + r. Rewritig the right-had side of this iequality ad the usig the AM-GM iequality o the parethesis yields (p + q + r = p + q + r + (p + q + r p + q + r + p q r = = p + q + r + pqr. What is ow left for us to show is that the iequality p q + pq + p r + pr + q r + qr p + q + r + pqr holds. Not much to show though, sice the iequality is Schur s iequality for r =. 6. We use the suggested otatio ad ( get ( + a ( + a... ( + a = + S + ( S +... + ( S. The we start( by showig ( that ( + S + S +... + S ( + S. With help from the biomial theorem we ( develop the right-had side of this iequality. The we get rid of the terms ad S, sice they exist o both sides of the iequality. We come up with ( the iequality ( ( ( ( ( S + S +...+ S S + S +...+ S. This iequality is true. As a matter of fact we fid that for all, ( =,,..., the :th term o the left-had side of the iequality is the :th term o the righthad side of the iequality. This we will show ow. We have iequalities: ( S ( S ( =,,...,. We divide both sides of all iequalities 6

by S (. The we tae the :th root of both sides to get the iequalities S ( =,,...,. These iequalities are obviously true; they follow directly from MacLauri s iequality. Now, what we have show is that ( + a ( + a... ( + a ( + (a a... a. Thus we have that ( + (a a... a. Taig the :th root of both sides of this iequality, the subtractig both sides by ad last taig the :th power of both sides of the iequality yields the desired iequality. A maybe easier solutio uses Hölder s iequality, as i problem 0. 7. We mae the suggested variable substitutio a = x, b = y, c = z. The we develop both sides of the iequality ad multiply both sides by x y z to get rid of the deomiators. We arrive at the iequality x 6 y z 0 x 5 y z, which sym sym is clearly true (Muirhead s iequality for a = 6, a =, a = 0, b = 5, b =, b =. 8. We start by multiplyig both sides of the iequality by 4(x + y(x + z(y + z(x + y + z to get rid of the deomiators. The we develop both sides of the iequality ad cacel terms that occur o both sides. Evetually we get the iequality 6xyz x y + x z + y x + y z + z x + z y, which also ca be expressed as xyz x yz 0. We see that this last iequality holds sym sym (Muirhead s iequality for a =, a =, a = 0, b = b = b =. 9. We mae the suggested substitutio. The we simplify the expressio usig well-ow trigoometric idetities. We get f(cos α, cos β = = cos α cos β+cos α cos β+cos β cos α ( cos α( cos β = = cos α cos β + cos α si β + cos β si α si α si β = = cos α cos β +cos α si β +cos β si α si α si β = cos(α+β+si(α+β = = ( cos(α + β + si(α + β = = (si π 4 cos(α + β + cos π 4 si(α + β = ( π si 4 + α + β. 7

The equality is attaied for α + β = π. Thus, for real umbers x, y, 4 the fuctio f(x, y has the maximum value. 40. We mae the suggested variable substitutio. The easy part of the problem is the to simplify the iequality we get from the substitutio. Usig well-ow trigoometric idetities we evetually arrive at the iequality ( si α + si β + si γ si α + si β + si γ. The coditio xy + yz + zx = becomes the coditio ta α ta β + ta β ta γ + ta γ ta α =. We multiply both sides of this equality by cos α cos β cos γ ad the do a little simplifyig. We ed up with the equality cos α + β + γ = 0. Solvig this equatio, we realize that the coditio α + β + γ = π i this case is equivalet to the coditio ta α ta β + ta β ta γ + ta γ ta α =. We bear i mid that we have α + β + γ = π ad the express the left-had side of ( i aother way, usig trigoometric idetities. We get si α + si β + si γ = si(α + β cos(α β + si γ = = si γ cos(α β + si γ cos γ = si γ(cos(α β + cos γ = = si γ(cos(α β cos(α + β = = si γ(cos α cos β+si α si β cos α cos β+si α si β = 4 si α si β si γ = = si α si β si γ cos α cos β cos γ. The we treat the right-had side of ( i a similar way. We get si α + si β + si γ = si α + β cos α β + si(α + β = = si α + β cos α β + si α + β cos α + β = = si α + β ( cos α β + cos α + β = = si α + β ( cos α cos β + si α si β + cos α cos β si α si β = = 4 si α + β cos α cos β ( π = 4 si γ cos α cos β = 4 cos α cos β cos γ. Thus our iequality to be prove becomes 8

si α si β si γ cos α cos β cos γ 4 cos α cos β cos γ, which, after divi- dig both sides of the iequality by 4 cos α cos β cos γ i tur becomes ( 8 si α si β si γ. Now we remid ourselves that we have 0 < α, β, γ < π. For f(x = si x we ote that f (x = si x 0 for 0 x π. We first apply the AM-GM iequality to the left-had side of ( ad the Jese s iequality (rememberig that Jese s iequality i this case is reversed to get 8 si α si β si γ ( 8 si α + si β + si γ ( 8 si α + β + γ ( π = 8 si =. 6 9

Additioal problems 4. Prove the iequality a b + c + b c + a + c a + b a + b + c 4. Let x, x,..., x be positive real umbers. Prove that x x xx...xx (x x...x x+x +...+x. 4. Prove that, for ay real umbers x, y, (x + y( xy ( + x ( + y. 44. (IMO 995 Let a, b ad c be positive real umbers such that abc =. Prove that a (b + c + b (a + c + c (a + b. 45. (Moldavia MO 996 Let a, b ad c be positive itegers such that a + b ab = c. Prove that (a c(b c 0. 46. (Irelad, 000 Let x ad y be oegative real umbers such that x+y =. Prove that x y (x + y. 47. (Thailad, 99 Let a, b, c ad d be positive real umbers such that ab + bc + cd + da =. Prove that a b + c + d + b c + d + a + c d + a + b + d a + b + c. 48. (Greece, 987 Let a, b ad c be positive real umbers. Show that for every a ( iteger, b + c + b c + a + c ( a + b + c a + b holds. 49. (Belarus, 99 Let x, x,..., x ( be oegative umbers such that x + x +... + x =. Prove that xi ( x i. 50. (Hugary, 985 Let a, b, c ad d be positive real umbers. Prove that a b + c + b c + d + c d + a + d a + b. 5. (Leigrad, 98 Let a, b ad c be real umbers such that 0 a, b, c.. 40

Prove that a( b( c + b( a( c + c( a( b + abc. 5. (Romaia, 004 Fid all positive real umbers a, b, c which satisfy 4(ab + bc + ca a + b + c (a + b + c. 5. (Romaia, 004 Let a, b ad c be real umbers such that a + b + c =. Prove that a + b + c abc 4. 54. (Estoia, 004 Let a, b ad c be positive real umbers such that a + b + c =. Prove that + ab + + bc + + ca. 55. (Austria, 004 Let a, b, c ad d be real umbers. Prove that a 6 + b 6 + c 6 + d 6 6abcd. 56. (Irelad, 004 Let a ad b be oegative real umbers. Prove that ( a(a + b + b a + b (a + b, with equality if ad oly if a = b. 57. (New Zealad, 004 Let x, x, y, y be positive real umbers. Prove that x + x (x + x. y y y + y 58. Let a, b ad c be real umbers. Prove that a + ( b + b + ( c + c + ( a. 59. Let a, b ad c be positive real umbers such that abc =. Prove that b + c + c + a + a + b a + b + c +. a b c 60. Let a, b ad c be positive real umbers. Prove that a (b + c + b (c + a + c (a + b 9 4(a + b + c. 6. (Juiors Bala MO 00 Let a, b ad c be positive real umbers such that abc =. Prove that a + b + c a b + c + b c + a + c a + b. 4

6. Let a, b ad c be positive real umbers such that abc. Prove that a b + b c + c a a + b + c. 6. (Juiors Bala MO 00 Let a, b ad c be positive real umbers. Prove that a b + b c + c a a b + b c + c a. 64. Let x, y ad z be positive real umbers such that x + y + z + xyz =. Prove that x + y + z. 65. (Juiors Bala MO 00 Let x, y ad z be real umbers such that + x x, y, z >. Prove that + y + z + + y + z + x + + z + x + y. 66. Let a, b ad c be positive real umbers such that a + b + c =. Prove that a + b b + c + b + c c + a + c + a a + b. 67. (IMO 987 Let x, y ad z be real umbers such that x + y + z =. Prove that x + y + z xyz +. 68. Let x, y, z ad α be positive real umbers such that xyz = ad α. x α Prove that y + z + yα z + x + zα x + y. 69. Let a, b, c ad d be positive real umbers. Prove that (a + b (b + c (c + d (d + a 6a b c d (a + b + c + d 4. 70. Let a, b ad c be positive real umbers such that a + b + c =. Prove that (a + b (b + c (c + a 8(a b + b c + c a. 4

Refereces Lee, Hojoo (00. Note o Muirhead s T heorem. Mitriović, Dragoslav S. (970. Aalytic Iequalities. Berli: Spriger-Verlag. Mitriović, Dragoslav S., Pečarić, J. E., Fi, A.M. (99. Classical ad New Iequalities i Aalysis. Dordrecht: Kluwer Academic Publishers. Persso, Are, Böiers, Lars Christer (990. Aalys i e variabel. Lud: Studetlitteratur. Rai, Robert A. (96. A Itroductio to Mathematical Aalysis. Oxford: Pergamo Press Limited. various problem collectios various upublished otes o algebraic iequalities URL http://plaetmath.org/ecyclopedia/proofofrearragemetiequality.html 4