The Australian Journal of Mathematical Analysis and Applications

The Australian Journal of Mathematical Analysis and Applications Volume 7, Issue, Article 11, pp. 1-14, 011 SOME HOMOGENEOUS CYCLIC INEQUALITIES OF THREE VARIABLES OF DEGREE THREE AND FOUR TETSUYA ANDO Received 19 May, 010; accepted 19 November, 010; published 30 March, 011. DEPARTMENT OF MATHEMATICS AND INFORMATICS, CHIBA UNIVERSITY, CHIBA 63-85, JAPAN ando@math.s.chiba-u.ac.jp ABSTRACT. We shall show that the three variable cubic inequality t (a 3 + b 3 + c 3 ) + (t 4 t)(ab + bc + ca ) (t 3 1)(a b + b c + c a) + (3t 4 6t 3 + 3t 6t + 3)abc holds for non-negative a, b, c, and for any real number t. We also show some similar three variable cyclic quartic inequalities. Key words and phrases: Homogeneous Inequality. 000 Mathematics Subject Classification. Primary 6D05. ISSN (electronic): 1449-5910 c 011 Austral Internet Publishing. All rights reserved.

TETSUYA ANDO 1. INTRODUCTION We denote cyclic or symmetric polynomials of three variables by S i,j,k := a i b j c k + b i c j a k + c i a j b k, S i,j := S i,j,0, S i := S i,0,0 = a i + b i + c i, T i,j,k := S i,j,k + S j,i,k, T i,j := T i,j,0 = S i,j + S j,i, T i := T i,0,0 = S i, U := 1 3 S 1,1,1 = 1 6 T 1,1,1 = abc. About symmetric polynomials, it is well known that Muirhead s inequalities S 3 = T 3 T,1 T 1,1,1 = 6U, S 4 = T 4 T 3,1 T, T,1,1 = US 1 hold for non-negative real numbers a, b, c. Schur s inequality and Cirtoaje s inequality ([3, p.73] ) (1.1) S 3 + 3U T,1, S 4 + (t + t)s, (t + 1)T 3,1 + (t 1)US 1 are rather non-trivial, here t is any real number. Any of the above quartic inequalities except T 3,1 T, hold even if some of a, b, c are negative. About cyclic polynomials, similar inequalities S 3 S,1 S 1,1,1 = 3U, S 4 S 3,1 S,1,1 = US 1, S 4 S, S,1,1 hold for non-negative a, b, c. Here, S 4 S 3,1 and S 4 S, S,1,1 are true for any real a, b, c. But S 3,1 S, does not hold even if a, b, c are non-negative. Schur type inequality S 3 + 3U S,1 does not also hold. But an weaker inequality 3 4 (1.) S,1 3 S 3 + ( 3 3 4 ) U 0.59134S 3 + 0.470866U holds for non-negative a, b, c. This will be proved as a corollary of the following Theorem. Theorem 1.1. For non-negative real numbers a, b, c and t, the following inequality holds: (1.3) t S 3 + (t 4 t)s,1 (t 3 1)S 1, + (3t 4 6t 3 + 3t 6t + 3)U. The equality holds if and only if a = b = c or a : b : c = t : 0 : 1 or any cyclic permutation thereof. This inequality will be proved in Section. Note that when t = 1 3 we obtain (1.), and when t = 1 + + 1 = 1.883035059135586416, we obtain S 3 + αs,1 (α + 1)S 1,, 16 + 13 1 here α = 5α 6α 3 = 0. =.48443533176585687519 which is a root of α 4 + α 3 Next, we consider quartic cyclic inequalities. It is well known the inequalities (1.4) S 4 + S 1,3 S 3,1, S 4 + (3t 1)S, 3tS 3,1 + 3t(t 1)US 1 hold for any real numbers t, a, b, c. Recently, Cirtoaje proved very nice inequality which includes (1.1) and (1.4)., Vol. 7, No., Art. 11, pp. 1-14, 011

SOME HOMOGENEOUS CYCLIC INEQUALITIES 3 Theorem 1.. ([4, Theorem.1] ). Let p, q, r be any real numbers. The cyclic inequality S 4 + rs, + (p + q r 1)US 1 ps 3,1 + qs 1,3 holds for any real numbers a, b, c if and only if In Section 3, we shall prove the following. 3(1 + r) p + pq + q. Theorem 1.3. (1) For any real numbers a, b, c and t such that t, the following inequality holds: (1.5) S 4 + t + 8 t(t ) S, + US 1 tt 3,1. 4 Here, the equality holds if and only if (a + b + c ) = t(ab + bc + ca). () For any non-negative real numbers a, b, c and t, the following inequality holds: (1.6) t 3 S 4 t (3 t 4 )S 3,1 + (3t 4 1)S 1,3 + (1 3t + t 3 3t 4 + t 6 )US 1. The equality holds if and only if a = b = c or a : b : c = t : 0 : 1 or any cyclic permutation thereof. As a special case of (1.6), when t = 1/ 4 3 we obain ( 4 4 ) 3 S 4 + 1 US 1 4 4 3 3 3 S 3,1. When t = 0.60755955811418156 is a root of t 6 3t 4 + t 3 3t + 1 = 0, we obtain S 4 + βs 1,3 (β + 1)S 3,1, here β = 1.3790744336539958016 which is a root of 4β 6 + 1β 5 48β 4 116β 3 + 4β + 84β + 9 = 0. Note that Theorem 1. does not included in Theorem 1.3, and Theorem 1.3(1) does not conflict to Theorem 1.: In this section, we shall show that (1.) S 4 + (p 1)S, + p( p)us 1 pt 3,1.. PROOF OF THEOREM 1.1 t (a 3 + b 3 + c 3 ) + (t 4 t)(ab + bc + ca ) Let u := a/c, v := b/c, and (t 3 1)(a b + b c + c a) + (3t 4 6t 3 + 3t 6t + 3)abc. x := S,1 S 3 = a b + b c + c a a 3 + b 3 + c 3 = u v + v + u u 3 + v 3 + 1, y := S 1, S 3 = ab + bc + ca a 3 + b 3 + c 3 = uv + v + u u 3 + v 3 + 1, z := U S 3 = abc a 3 + b 3 + c 3 = uv u 3 + v 3 + 1., Vol. 7, No., Art. 11, pp. 1-14, 011

4 TETSUYA ANDO Eliminate u, v from x, y, z, we obtain the relation (.1) x 3 + y 3 + 9z 3 6xyz xy + 3z + z = 0, here x 0, y 0, z 0. (.1) determines the rational algebraic surface S as the following figure. Let C be the intersection of S and the (x, y)-plane. C is the rational cubic curve defined by x 3 + y 3 xy = 0. The part of C in x 0, y 0 is a convex closed curve. Note that any point Q on C can be represented as x = u u 3 + 1 =: f(u), y = u u 3 + 1 =: g(u) (u 0). S have a rational double point of the type A 1 at P = (1, 1, 1/3) as the unique singularity. In fact, let Then (.1) is equivalent to X := 1 x, Y := 1 y, Z := 1 3 x. 3X + 3Y + 1Z 3XY 6Y Z 6XZ X 3 Y 3 9Z 3 + 6XY Z = 0. With parameters A := X/Z and B := Y/Z, we can represent X, Y, Z as X = 3(4A A AB + A 3 A B + AB ) 9 6AB + A 3 + B 3, Y = 3(4B B AB + A B AB + B 3 ) 9 6AB + A 3 + B 3, Z = 3(4 A B + A + B AB) 9 6AB + A 3 + B 3. This means that every line passing though P meet again with S at only one point. We shall call such surface to be a concave cone. Let Q := ( f(u), g(u) ) be a point on C, and let R 3 + be the first quadrant of (x, y, z)-space defined by x 0, y 0, z 0. Let H be the plane which tangents to C at Q, and which passes, Vol. 7, No., Art. 11, pp. 1-14, 011

SOME HOMOGENEOUS CYCLIC INEQUALITIES 5 through P. Since P Q S = {P, Q}, and since C is a convex closed curve, we conclude that H S R 3 + = {P, Q}. The equation of H is x y z 1 1 1 1 1 0 = 3 f(u) g(u) 0 1 f(u) + d f(u) g(u) + d g(u) 0 1 du du = u (u 4 u)x + (u 3 1)y + (3u 4 6u 3 + 3u 6u + 3)z. 3(u 3 + 1) The numerator of the above gives (1.), after we observe which side of H there is S. 3. PROOF OF THEOREM 1. Proof of (1.5). Let a, b, c be any real numbers, u := a/c, v := b/c, and let x := T 3,1 S 4 = u3 v + v 3 u + u 3 + v 3 + u + v u 4 + v 4 + 1 =: p(u, v), y := S, = u v + u + v =: q(u, v), S 4 u 4 + v 4 + 1 z := US 1 uv(u + v + 1) = =: r(u, v). S 4 u 4 + v 4 + 1 These parametrization determines a rational quadric surface S defined by (x + z) (y + 1)(y + z) = 0. Note that S is an elliptic cone with vertex ( 1/4, 1/, 1/4). The above figure is the part of S determined by 0 u 1, 0 v 1. This looks like a triangular sail of a yacht. Note that since the parametrization is not one to one, the (x, y, z) can move limited area of S. Now we calculate the tangent plane of S. Let A 0 := (u + v + uv) A 1 := (u + v + uv)(1 + u + v ) A := 1 + 4u + 4uv + 4v + 4u v + 4uv + u 4 + 4u v + v 4 A 3 := (1 + u + v )(1 u v + u uv + v )., Vol. 7, No., Art. 11, pp. 1-14, 011

6 TETSUYA ANDO Since x y z 1 p(u, v) q(u, v) r(u, v) 1 p(u, v) + p(u, v) q(u, v) + q(u, v) r(u, v) + r(u, v) 1 u u u p(u, v) + p(u, v) q(u, v) + q(u, v) r(u, v) + r(u, v) 1 v v v = (u 1)(v 1)(u v)(u + v + 1) (1 + u 4 + v 4 ) 3 ( A0 A 1 x + A y + A 3 z ), the tangent plane H of S at ( p(u, v), q(u, v), r(u, v) ) is given by (3.1) A 0 A 1 x + A y + A 3 z = 0. By the way, since 4 A A 0 = ( A1 A 0 we can represent (3.1) by the parameter as ) + 8, A 3 A 0 = t := A 1 = (1 + u + v ) A 0 u + v + uv (3.) H : 1 tx + t + 8 4 ( A1 A 0 ) A 1 A 0, = (a + b + c ) ab + bc + ca y + t t z = 0. Sinc S is a cone, S is contained in a half space whose boundary is H. But since (1 + u + v ) (u + v + uv) = (u v) + (u 1) + (v 1) 0, t can only take any real numbers such that t. Thus we complete the proof. Note that H does not pass through (, 1, 1), unless t =. Proof of (1.6). We may assume that a, b, c be real numbers. Let u := a/c, v := b/c, and let x := S 3,1 S 4 y := S 1,3 S 4 z := US 1 S 4 = = u3 v + v 3 + u u 4 + v 4 + 1 = uv3 + u 3 + v u 4 + v 4 + 1 uv(u + v + 1) u 4 + v 4 + 1 =: p(u, v), =: q(u, v), =: r(u, v). These parametrization determines a rational quadric surface S defined by x 4 + y 4 + 3z 4 + x y + 4y z + 4x z 4xz 3 4xyz 4yz 3 + 5z 3 4xz 4yz + x z 4xyz + y z xy xz yz + 4z + z = 0., Vol. 7, No., Art. 11, pp. 1-14, 011

SOME HOMOGENEOUS CYCLIC INEQUALITIES 7 S has a rational double point at P = (1, 1, 1). Let C be the intersection of S and the (x, y)-plane. C is a rational curve with a parametrization (3.3) x = u u 4 + 1 =: p 0(u), y = u3 u 4 + 1 =: q 0(u). The part of C in x 0, y 0 is a convex closed curve. Let Q be the point on C defined by (3.3), and let H be the tangent plane of C at Q which passes through P. The equation of H is obtained by x y z 1 1 1 1 1 0 = p 0 (u) q 0 (u) 0 1 p 0 (u) + d p du 0(u) q 0 (u) + d q du 0(u) 0 1 = u (3 u 4 )x (1 3u 4 )y + (1 3u + u 3 3u 4 + u 6 )z u 3. (1 + u 4 ) If we show that H S R 3 + = {P, Q}, we know that S must be contained in the half space defined by u (3 u 4 )x (1 3u 4 )y + (1 3u + u 3 3u 4 + u 6 )z u 3 0. Then we obtain (1.6). Let s show that H S R 3 + = {P, Q}, Note that the intersection number P Q and S at P is not less than. Since S is quartic, and since S is closed surface, we have P Q S = {P, P, Q, Q }, here Q is the third intersection of P Q and S. It is enough to show that the z-coordinate of Q is not positive. Assume that the z-coordinate of Q is positive. Then, we can find a point Q S R 3 + such that the line P Q tangents to S. Let H be the tangent plane of S at Q = ( p(u, v), q(u, v), r(u, v) ). Let A 0 := u 4 v 4 + u 6 v + u 5 v u 4 v 3 u 3 v 4 + u v 5 + uv 6 + u 5 v u 4 v + 3u 3 v 3 u v 4 + uv 5 u 4 v + 3u 3 v + 3u v 3 uv 4 u 4 u 3 v u v uv 3 v 4 + u v + uv + uv, A 1 := 3u 5 v 3 uv 7 u 7 u 6 v + u 5 v + u 4 v 3 3u 3 v 4 + u v 5 uv 6 + u 5 v u 4 v u v 4 + uv 5 3u 4 v + uv 4 + 3v 5 + u 3 v u v 3uv 3 + 3u 3 + u v + uv uv v,, Vol. 7, No., Art. 11, pp. 1-14, 011

8 TETSUYA ANDO A := u 7 v + 3u 3 v 5 u 6 v + u 5 v 3u 4 v 3 + u 3 v 4 + u v 5 uv 6 + v 7 + u 5 v u 4 v u v 4 + uv 5 + 3u 5 + u 4 v 3uv 4 3u 3 v u v + uv 3 + u v + uv + 3v 3 uv u, A 3 := u 8 u 7 v + u 6 v u 5 v 3 u 4 v 4 u 3 v 5 + u v 6 uv 7 + v 8 u 7 + u 4 v 3 + u 6 v + u 3 v 4 + uv 6 v 7 + u 6 9u 4 v 9u v 4 + 8u 3 v 3 + v 6 u 5 + u 4 v + 8u 3 v + 8u v 3 + uv 4 v 5 u 4 + u 3 v 9u v + uv 3 v 4 u 3 v 3 + u + uv + v u v + 1. Since x y z 1 p(u, v) q(u, v) r(u, v) 1 p(u, v) + p(u, v) q(u, v) + q(u, v) r(u, v) + r(u, v) 1 u u u p(u, v) + p(u, v) q(u, v) + q(u, v) r(u, v) + r(u, v) 1 v v v = 1 + u + v ( A0 + A (1 + u 4 + v 4 ) 3 1 x + A y + A 3 z ), the equation of the tangent plane of S at Q = ( p(u, v), q(u, v), r(u, v) ) is (3.4) A 0 + A 1 x + A y + A 3 z = 0. Since H passes through P = (1, 1, 1), we have a relation A 0 + A 1 + A + A 3 = (1 + u + v)(1 u v + u + v uv) (1 u v u v + 3uv + u 3 u v uv + v 3 ) = 0. Note that (1 u v + u + v uv) = (u v) + (u 1) + (v 1) > 0, if Q P. If u + v + 1 = 0, then Q = ( 1/, 1/, 0). Thus we consider a relation 1 u v u v + 3uv + u 3 u v uv + v 3 = 0. This equation determines a rational cubic curve Γ in (u, v)-plane with a singularity at (1, 1)., Vol. 7, No., Art. 11, pp. 1-14, 011

SOME HOMOGENEOUS CYCLIC INEQUALITIES 9 Let t := (1 v)/(1 u). Then Γ have a parametrization (3.5) u = t (t ) 1 t t + t, v = 1 t 3 1 t t + t. 3 Thus we see that { (u, v) Γ u > 0, v > 0 } = {(1, 1)}. Note that (u, v) = (1, 1), if t 3 = 1 and t 1. Thus u 0 or v 0, and r(u, v) 0. This is a contradiction. By the above observation, we obtain another inequality. Assume that a, b, c are any real numbers. Substitute (3.5) for (3.4), and multiply then we obtain (1 t) 16 (1 + t) 8 (1 t + t ) 3 ( 1t + 18t + 0t 3 51t 4 48t 5 + 144t 6 48t 7 51t 8 + 0t 9 + 18t 10 1t 11 + t 1) 1, (1 6t + 6t + 8t 3 9t 4 + t 6 )x + (1 9t + 8t 3 + 6t 4 6t 5 + t 6 )y = (1 3t 3t + 11t 3 3t 4 3t 5 + t 6 )z + (1 3t + 5t 3 3t 5 + t 6 ). Since S is contained in a half space divided by H, thus we have the following: (1 3t + 5t 3 3t 5 + t 6 )S 4 + (1 3t + t 3 )(1 3t + t 3 )US 1 But this is a special case of Theorem 1.. (1 6t + 6t + 8t 3 9t 4 + t 6 )S 3,1 + (1 9t + 8t 3 + 6t 4 6t 5 + t 6 )S 1,3. It is haeder to obtain inequalities about S 4, S,, S 3,1 and S 1,3. Let a, b, c be any real numbers, u := a/c, v := b/c, and let x := S 3,1 S 4 = u3 v + v 3 + u u 4 + v 4 + 1 =: p(u, v),, Vol. 7, No., Art. 11, pp. 1-14, 011

10 TETSUYA ANDO uv 3 + u3 + v S1,3 = 4 =: q(u, v), S4 u + v4 + 1 S, u v + u + v z := = =: r(u, v). S4 u4 + v 4 + 1 y := This determines a rational quartic surface S as the following figure. S has an unique singular point at P = (1, 1, 1). Note that S R3+ is not concave cone. Thus we can not use the above method. Anyway, we calculate the equation of the tangent plane H of S at Q = p(u, v), q(u,, v), r(u, v). Let B0 := u6 v u5 v 3 + u4 v 4 u3 v 5 + u v 6 u5 v u4 v 3 u3 v 4 u v 5 + u6 u5 v + 5u4 v + 5u v 4 uv 5 + v 6 u5 u4 v uv 4 v 5 + u4 u3 v + 5u v uv 3 + v 4 u3 u v uv v 3 + u + v, B1 := (u + v + 1) u5 v u4 v u4 v u3 v + u v 3 uv 4 + v 5 + u3 v + 3u v uv 3 v 4 u v + uv u uv + u, B := (u + v + 1) u v 4 + uv 5 + u5 u4 v + u3 v u v 3 uv 4 u4 u3 v + 3u v + uv 3 + u v uv uv v + v, B3 := u8 u7 v + u6 v u5 v 3 u4 v 4 u3 v 5 + u v 6 uv 7 + v 8 u7 + u6 v + u4 v 3 + u3 v 4 + uv 6 v 7 + u6 9u4 v + 8u3 v 3 9u v 4 + v 6 u5 + u4 v + 8u3 v + 8u v 3 + uv 4 v 5 u4 + u3 v 9u v + uv 3 v 4 u3 v 3 + u + uv + v u v + 1., Vol. 7, No., Art. 11, pp. 1-14, 011

SOME HOMOGENEOUS CYCLIC INEQUALITIES 11 Since H is defined by x y z 1 p(u, v) q(u, v) r(u, v) 1 p(u, v) + p(u, v) q(u, v) + q(u, v) r(u, v) + r(u, v) 1 u u u p(u, v) + p(u, v) q(u, v) + q(u, v) r(u, v) + r(u, v) 1 v v v = 1 + u + v ( B0 + B (1 + u 4 + v 4 ) 3 1 x + B y + B 3 z ), (3.6) B 0 + B 1 x + B y + B 3 z = 0. We assume that H passes through P = (1, 1, 1). Then we have a relation B 0 + B 1 + B + B 3 = (1 u v + u uv + v ) ( u 4 u 3 v u v uv 3 + v 4 u 3 + u v + uv v 3 u + uv v u v + 1 ) = 0 Since 1 u v + u uv + v > 0 if (x, y, z) P, we only consider the case u 4 u 3 v u v uv 3 + v 4 u 3 + u v + uv v 3 u + uv v u v + 1 = 0 This define a rational quartic curve which can be parameterized by t := (1 v)/(1 u) as (3.7) u = 1 + t + t 3t 3 + t 4 1 t t t 3 + t 4, v = 1 3t + t + t 3 t 4 1 t t t 3 + t 4. Substitute (3.7) for (3.6), and multiply we obtain (1 t t t 3 + t 4 ) 8 ( 5t + t + t 3 5t 4 + t 5 ) 4 (1 3t + 5t 3 3t 5 + t 6 ), (1 3t + 5t 3 3t 5 + t 6 ) + 3t(1 t)(1 t t + t 3 )x 3t(1 t)(1 t t + t 3 )y (1 3t + t 3 )(1 3t + t 3 )z = 0. Since S is contained in a half space divided by H, we have the following: (1 3t + 5t 3 3t 5 + t 6 )S 4 + 3t(1 t)(1 t t + t 3 )S 3,1 (1.7) 3t(1 t)(1 t t + t 3 )S 1,3 + (1 3t + t 3 )(1 3t + t 3 )S,. But this is also a special case of Theorem 1.. Some reader may want to obtain some inequalities with S 4, S 3,1, S,, US 1 by the same way. Let u := a/c, v := b/c and x := S 3,1 S 3 = u3 v + v 3 + u u 4 + v 4 + 1, y := S, S 3, z := US 1 S 4. This determines a rational quartic surface S as the following figure., Vol. 7, No., Art. 11, pp. 1-14, 011

1 TETSUYA ANDO The singularities of S are not only the point P = (1, 1, 1) but also a curve as in the figure. Moreover S R 3 + is not concave cone. This observation will not succeed. But see Remark 4.1. 4. APPENDIX: NOTE FOR THE ELEMENTARY SYMMETRIC POLYNOMIALS For the elementary symmetric polynomials S 1 = a + b + c, S 1,1 = ab + bc + ca and U = abc, the following theorem holds. Theorem 4.1. For non-negative real numbres a, b, c, (4.1) 4S 3 1U S 1S 1,1 + 4S 3 1,1 18S 1 S 1,1 U + 7U 0 holds. Proof. Let x := U S 3 1 have, y := S 1,1, and t := c. Since S S1 1,1 = ab + c(a + b) = U S 1 c + c(s 1 c), we (4.) x ty + t t 3 = 0. Let L t be the line in (x, y)-plane defined by (4.). The envelope of the family L t is the curve C defined by Note that L t tangents to C at 4y 3 y 18xy + 7x + 4x = 0 (x, y) = ( t (1 t), t( 3t) ) =: ( p(t), q(t) ), and this is a parametrization of C. C has a singular point at P = ( 1 7, 1 ) when t = 1/3. 3, Vol. 7, No., Art. 11, pp. 1-14, 011

S OME HOMOGENEOUS CYCLIC 13 INEQUALITIES Let C1 be the part of C with 0 t 1/3, and C be the part with 1/3 t 1/. We shall prove that when a, b, c vary all non-negative real numbers, (x, y) varies all the points of the closed domain D defined by 4y 3 y 18xy + 7x + 4x 0, x 0, y 0. We may assume c a, c b and a + b + c = 1. Then 0 t 1/3. Thus the point (p(t), q(t)) moves all the point on C1. Since C1 is concave, Lt exists above C1. Since a+b 1 t x = tab t =t, x varies 0 x t(1 t) /4. By (4.), when x = t(1 t) /4, then y = 1 + t 3t. This point (x, y) lies on C. Since C is convex, we have y S3 + 3U T,1. 9x + 1. Thus S13 + 9U 4S1 S1,1. This is equivalent to 4 Remark 4.1. We obtained a more strict inequality than (1.6), during this article is under publishing process. We only present its statement: For non-negative real numbers a, b, c and s, and for t 1, the following inequality holds: t 1 (4.3) S4 s S3,1 st S1,3 + s + t S, s s s t 1 + 1 (s 1) 1 + + U S1 0. s s The equality holds if and only if a = b = c or a : b : c = 0 : s : 1 or any cyclic permutation thereof. Conversely, if t < 1 and s > 0, there exists non-negative a, b, c for which (4.3) does not hold., Vol. 7, No., Art. 11, pp. 1-14, 011

14 TETSUYA ANDO Remark 4.. We also announce the following: Let F t := t S 3 + (t 4 t)s,1 (t 3 1)S 1, (3t 4 6t 3 + 3t 6t + 3)U, F := S,1 3U. If f(a, b, c) is a cubic homogeneous cyclic polynomial such that f(a, b, c) 0 for any a 0, b 0, c 0, and that f(1, 1, 1) = 0, then there exists λ i 0 and s i [0, ] (i = 1,, 3) such that f = λ 1 F s1 + λ F s + λ 3 F s3. REFERENCES [1] W. P. BARTH, K. HULEK, C. A. M. PETERS and A. V. DE VEN, Compact Complex Surfaces, Second Enlarged Edition, Springer, (003). [] A. BEAUVILLE, Complex Algebraic Surfaces, Second Edition, LMSST34, London Math. Soc. (1996). [3] V. CÎRTOAJE, Algebraic Inequalities Old and New Methods, GIL Publishing House, Romania, (006). [4] V. CÎRTOAJE, On the Cyclic Homogeneous Polynomial Inequalities of Degree Four, J. Inequal. Pure and Appl. Math., 10(3) (009), Art. 67, pp. 1-10. [5] P. K. HUNG, Secrets in Inequalities, Volume Advanced Inequalities, GIL Publishing House, Romania, (008)., Vol. 7, No., Art. 11, pp. 1-14, 011