MATH 289 WINTER 2011 PROBLEM SET 3: INEQUALITIES

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1 MATH 89 WINTER 0 PROBLEM SET 3: INEQUALITIES. Elemetary iequalities Perhaps the most fudametal iequality for real umbers is x 0, x R. Usig this iequality oe ca deduce may more iequalities. For example, if we take x = a b with a, b R we obtai: a ab + b = (a b) 0. It follows that a + b ab. This iequality is iterestig by itself. If we ow substitute a = y ad b = z we obtai y + z yz. wheever y, z are oegative real umbers. Substitutio is a very useful method for provig iequalities. Example. Prove that a + b + c ab + ac + bc for all a, b, c R. Also prove that equality holds if ad oly if a = b = c. Proof. We have a + b + c ab ac bc = ((a b) + (b c) + (c a) ) 0 ad it is ow obvious that equality holds if ad oly if a = b = c = 0. Aother obvious but importat iequality is: xy 0, if x, y R ad x 0 ad y 0. This ca be used i may ways. For example if 0 x the x x because x x = x( x) 0 ad both x ad x are oegative. Example. Suppose that x, x,...,x are real umbers such that 0 x i for all i. Prove that x + x + + x x x + x x 3 + x 3 x x x. Whe do we have equality?

2 Proof. The iequality is obvious because it is equivalet to If we have equality the x ( x ) + x ( x 3 ) + + x ( x ) 0. x = 0 or x =, x = 0 or x 3 =,..., x = 0 or x =. If x 0 the x = ad i particular x 0. From this it follows that x 3 =. But the x 3 0, so x 4 =, etc. This way we see that x = x 3 = x 4 = = x = x =. I a similar way we see that if x i 0 for some i, the x = x = x 3 = = x =. The oly other case where equality holds is whe x = x = = x = 0. Makig the right substitutios ca be very helpful as the followig example shows. Example 3. Suppose that a, a,..., a are real umbers such that a i for all i. Prove the iequality ( + a )( + a ) ( + a ) + ( + a + a + + a ). Let us write a i = x i +. The x i 0 for all i. It is easier to deal with the iequality x i 0 tha with the iequality a i. The iequality trasforms to ( + x )( + x ) ( + x ) + (x + x + + x + ( + )) = = + + (x + x + + x ). This iequality follows already if we oly look at the costat ad liear part of the lefthadside: ( + x )( + x ) ( + x ) + (x + x + + x ) + + (x + + x ). because + =. Covexity +. Let f be a real-valued fuctio o a iterval I R. Now f is said to be covex if f(ta + ( t)b) tf(a) + ( t)f(b) for all t [0, ] ad all a, b I (the chord betwee (a, f(a)) ad (b, f(b)) lies above the graph of f). The fuctio f is said to be cocave if f(ta + ( t)b) tf(a) + ( t)f(b) for all t [0, ] ad all a, b I (the chord betwee (a, f(a)) ad (b, f(b)) lies below the graph of f).

3 cocave covex (You may well be used to a differet termiology, for example cocave up ad cocave dow istead of covex ad cocave.) Theorem 4. Suppose that f is a real-valued fuctio o I R, x, x,...,x I, ad t, t,...,t [0, ] with t + t + + t =. If f is covex, the () f(t x + t x + + t x ) t f(x ) + t f(x ) + + t f(x ). If f is cocave, the () f(t x + t x + + t x ) t f(x ) + t f(x ) + + t f(x ). Proof. Suppose that f is covex. We will prove the statemet by iductio o, the case = beig trivial. Suppose that we already have prove that f(t x + t x + + t x ) t f(x ) + t f(x ) + + t f(x ). for all x, x,...,x I ad all t, t,...,t [0, ] with t + t + + t =. Suppose ow that x, x,...,x + I ad t,...,t + [0, ] with t +t + +t + =. Defie s i = t i /( t + ) for i =,,...,. Note that s + s + + s =. Take a = s x + s x + + s x, b = x + ad t = t +. From the defiitio of covexity ad the iductio hypothesis follows that f(t x + + t + x + ) = f(ta + ( t)b) tf(a) + ( t)f(b) = = ( t + )f(s x + + s x ) + t + f(x + ) ( t + )(s f(x ) + s f(x ) + + s f(x )) + t + f(x + ) = = t f(x ) + + t + f(x + ). To prove the secod statemet, observe that f is cocave if ad oly if f is covex. The apply the first statemet to f. I particular the case t = t = = t = / is iterestig. Corollary 5. If f is covex o I, the f ( x + x + + x ) f(x ) + + f(x ) for all x,...,x I. If f is cocave o I, the f ( x + x + + x ) f(x ) + + f(x ) for all x,...,x I. 3

4 Theorem 6. Suppose that f is a real-valued fuctio o a iterval I R with a secod derivative. If f (x) 0 for all x I, the f is covex. If f (x) 0 for all x I, the f is cocave. (The coverse of these statemets are also true). Proof. If f (x) 0 for all x I the f (x) is weakly icreasig o the iterval I. Suppose that a, b I ad t [0, ]. Defie c = ta + ( t)b. By the Mea Value Theorem, there exist α (a, c) ad β (c, b) such that f f(c) f(a) (α) = ad f (β) = c a Sice α < β ad f is weakly icreasig, we have Multiplyig out gives f(ta + ( t)b) f(a) ( t)(b a) f (β) = f(b) f(c) b c = = f(c) f(a) c a f(b) f(c). b c = f (α) f(b) f(ta + ( t)b) t(b a) f(ta + ( t)b) tf(a) + ( t)f(b). This shows that f is covex. The secod statemet follows from the first statemet, applied to f. Example 7. Suppose that α, β, γ are the agles of a triagle. Prove that si(α) + si(β) + si(γ) 3 3 Proof. The fuctio si(x) is cocave o the iterval [0, π], because its secod derivative is si(x) 0. Thus we have si(α) + si(β) + si(γ) 3 si ( α + β + γ) = si( 3 3 π π) =. 3. Arithmetics, Geometric ad Harmoic mea Theorem 8. Let x, x, x 3,...,x > 0. We defie the Arithmetic Mea by A(x, x,...,x ) = x + x + + x, the Geometric Mea by G(x, x,...,x ) = x x x ad the Harmoic Mea by H(x, x,...,x ) = x + x + +. x The we have H(x,...,x ) G(x,..., x ) A(x,...,x ). 4

5 Proof. Let f(x) = log(x). The f (x) = /x < 0 for x > 0 so f is cocave o the iterval (0, ). It follows that log ( x + x + + x ) log(x ) + log(x ) + + log(x ). Applyig the expoetial fuctio (which is a icreasig fuctio) to both sides yields If we ow take y i = x i the we get x + x + + x x x x. Takig the reciprocal yields y + y + + y y y y. y + y + + y y y. y Example 9. Suppose that x, x,..., x are positive real umbers. Prove that x + x + + x + x. x x 3 x x Proof. Put y i = x i /x i+ for all i. We assume that the idex is cyclic, so that x + = x. Comparig the arithmetic ad geometric average gives: y + y + + y y y y =. 4. The Schwarz Iequality Aother importat iequality is the Schwarz iequality. For vectors x = (x,...,x ) ad y = (y,...,y ) i R oe defies x y = x y + + x y. Note that x y = y x, (x+y) z = x z +y z ad (tx) y = t(x y) for t R ad x, y, z R. The orm of the vector x is defied by x = x x = x + + x. Theorem 0. Suppose that x = (x, x,...,x ), y = (y, y,...,y ) R, the x y + + x y x + + x y + + y or i short form: x y x y. 5

6 Proof. For ay vector a a 0. I particular, if we take a = x + ty we get (x + ty) (x + ty) = x x + t(x y) + t (y y) 0 for all t 0. Viewed as a quadratic polyomial i t, this polyomial has a opositive discrimiat. The discrimiat is I particular we have ad takig square roots gives us 4(x y) 4(x x)(y y) 0 (x y) (x x)(y y) x y x x y y = x y. The Schwarz iequality is importat i Euclidea geometry i dimesio, 3 or higher. I particular, oe ofte defies the agle φ betwee two vectors x, y by cos(ϕ) = x y x y, 0 ϕ π. The Schwarz iequality tells us that this defiitio makes sese, sice the righthad side has absolute value at most. 5. The triagle iequality Aother famous geometric iequality is the triagle iequality. If a, b, c are the legths of the sides of a triagle, the a + b c (ad also a + c b ad b + c a). 6. Oe more useful iequality Theorem. Suppose that x, x,...,x, y, y,...,y are real umbers such that x x x ad y y y. Suppose that z, z,..., z are the same as y, y,...,y, but possibly i a differet order. The we have x y + x y + + x y x z + x z + + x z x y + x y + + x y. Proof. Suppose that z, z,...,z is a rearragemet of y, y,...,y. Let m be the umber of displacemets of the sequece z, z,..., z, so m is the umber of pairs (i, j) with i < j ad z i > z j. We prove the right iequality by iductio o m. If m = 0 the z i = y i for all i ad we have iequality. Suppose m > 0. The z i > z i+ for some i. Note that the sequece z, z,...,z i, z i+, z i, z i+,...,z (exchage z i ad z i+ ) has oly m displacemets, so by iductio so We have x z + x z + x i z i+ + x i+ z i + + x z x y + x y + + x z. (x i+ x i )(z i z i+ ) 0, x i z i + x i+ z i+ x i z i+ + x i+ z i 6

7 ad x z + x z + + x i z i + x i+ z i+ + + x z x z + x z + + x i z i+ + x i+ z i + + x z x y + x y + + x y. The left iequality i the Theorem follows from the right iequality. Note that y y + y ad that z, z,..., z is a rearragemet of y, y,..., y. So we have x ( z ) + x ( z ) + + x ( z ) x ( y ) + x ( y ) + + x ( y ). 7. Exercises Exercise. ** Use the iequality x+y xy repeatedly to prove x + y + z + w 4 xyzw 4 for all x, y, z, w 0. Exercise. ** Prove that for all positive itegers. x + x + + x Exercise 3. * If x y z ad y > 0, prove that x + z y xz y i<j Exercise 4. ** For oegative real u,...,u, prove that ( u i ) 3 u 3 i. (use that x 3 is covex for x 0). i= Exercise 5. *** Suppose that p, p,..., p are positive real umbers such that i= p i =. Prove that p i log p i log. i= (This iequality comes from iformatio theory. ) Exercise 6. ** For positive real a, b, c prove that Exercise 7. *** Let Prove that i= b 3 c 3 + c 3 a 3 + a 3 b 3 3a b c. s = x i x j (( + ) ) s /( ). 7

8 Exercise 8. **** Prove the Hölder iequality: If /p + /q = ad x, y R the where x p = ( x p + x p + + x p ) /p. x y x p y q Exercise 9. * Let Q be a covex quadrilateral (i.e., the diagoals lie iside the figure). Let S be the sum of the legths of the diagoals ad let P be the perimeter. Prove P < S < P. Exercise 0. ** Suppose that we have a triagle with sides a, b, c such that for every positive iteger there exists a triagle with sides a, b ad c. Prove that the triagle must be isosceles. Exercise. ** Suppose that x, x,...,x are positive real umbers. Prove that Exercise. *** Prove that for all positive real umbers a, b, c. x + x + + x + x x + x + + x. x x 3 x x a a b b c c a b b c c a Exercise 3. * Prove that x x + x + x x + x x x + x + x x + x Exercise 4. **** Prove or disprove: If x ad y are real umbers with y 0 ad y(y+) (x + ), the y(y ) x. Exercise 5. **** Let a, b, c be positive real umbers such that abc =. Prove that = a 3 (b + c) + b 3 (c + a) + c 3 (a + b) 3. Exercise 6. **** Let p, p,...,p be ay poits o the sphere {(x, y, z) x + y + z = }. Prove that the sum of the squares of the distaces betwee them is at most. Exercise 7 (USSR Mathematics Olympiad). ****** Suppose that x, x,...,x are positive real umbers. Prove that x + x + + x x + x 3 x 3 + x 4 x + x 4 (idices go cyclic). 8