Some strong sufficient conditions for cyclic homogeneous polynomial inequalities of degree four in nonnegative variables

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1 Available online at J. Nonlinear Sci. Appl. 6 (013), Research Article Some strong sufficient conditions for cyclic homogeneous polynomial inequalities of degree four in nonnegative variables Yuanzhe Zhou a, Vasile Cirtoaje b, a The School of Physics Technology at Wuhan University, China. b Department of Automatic Control Computers, University of Ploiesti, Ploiesti, Romania. Dedicated to the memory of Professor Viorel Radu Communicated by Adrian Petrusel Abstract We establish some strong sufficient conditions that the inequality f 4 (x, y, z) 0 holds for all nonnegative real numbers x, y, z, where f 4 (x, y, z) is a cyclic homogeneous polynomial of degree four. In addition, in the case f 4 (1, 1, 1) = 0 also in the case when the inequality f 4 (x, y, z) 0 does not hold for all real numbers x, y, z, we conjecture that the proposed sufficient conditions are also necessary that f 4 (x, y, z) 0 for all nonnegative real numbers x, y, z. Several applications are given to show the effectiveness of the proposed methods. Keywords: Cyclic homogeneous polynomial; strong sufficient conditions; necessary sufficient conditions; nonnegative real variables. 010 MSC: Primary 6D Introduction Consider first the third degree cyclic homogeneous polynomial f 3 (x, y, z) = x 3 + Bxyz + C x y + D xy, where B, C, D are real constants, denotes a cyclic sum over x, y z. In [6], Pham Kim Hung gives the necessary sufficient conditions that f 3 (x, y, z) 0 for any nonnegative real numbers x, y, z. Corresponding author address: vcirtoaje@upg-ploiesti.ro (Vasile Cirtoaje) Received

2 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), Theorem 1.1. The cyclic inequality f 3 (x, y, z) 0 holds for all nonnegative real numbers x, y, z if only if f 3 (1, 1, 1) 0 for all nonnegative real x. f 3 (x, 1, 0) 0 Consider now the fourth degree cyclic homogeneous polynomial f 4 (x, y, z) = x 4 + A x y + Bxyz x + C x 3 y + D xy 3, where A, B, C, D are real constants. The following two theorems in [4] express the necessary sufficient conditions that f 4 (x, y, z) 0 for any real numbers x, y, z. Theorem 1.. The cyclic inequality f 4 (x, y, z) 0 holds for all real numbers x, y, z if only if for all real t, where k [0, 1] is a root of the equation f 4 (t + k, k + 1, kt + 1) 0 (C D)k 3 + (A B C + D 4)k (A B + C D 4)k + C D = 0. Theorem 1.3. The cyclic inequality f 4 (x, y, z) 0 holds for all real numbers x, y, z if only if g 4 (t) 0 for all t 0, where g 4 (t) = 3( + A C D)t 4 F t 3 + 3(4 B + C + D)t A + B + C + D, F = 7(C D) + E, E = 8 4A + B C D. In the particular case f 4 (1, 1, 1) = 0, from Theorem 1.3 we get the following corollary (see [1] [3]): Corollary 1.4. If 1 + A + B + C + D = 0, then the cyclic inequality f 4 (x, y, z) 0 holds for all real numbers x, y, z if only if 3(1 + A) C + CD + D. The following propositions in [4] give the equality cases of the inequality f 4 (x, y, z) 0 in Theorem 1. Theorem 1.3, respectively. Proposition 1.5. The cyclic inequality f 4 (x, y, z) 0 in Theorem 1. becomes an equality if x t + k = y k + 1 = z kt + 1 (or any cyclic permutation), where k (0, 1] is a root of the equation (C D)k 3 + (A B C + D 4)k (A B + C D 4)k + C D = 0 t R is a root of the equation f 4 (t + k, k + 1, kt + 1) = 0.

3 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), Proposition 1.6. For F > 0, the cyclic inequality f 4 (x, y, z) 0 in Theorem 1.3 becomes an equality if only if x, y, z satisfy (C D)(x + y + z)(x y)(y z)(z x) 0 are proportional to the real roots w 1, w w 3 of the equation w 3 3w + 3(1 t )w + E F t3 + 3t 1 = 0, where t is any double nonnegative real root of the polynomial g 4 (t). The following theorem in [5] expresses some strong sufficient conditions that the inequality f 4 (x, y, z) 0 holds for any real numbers x, y, z. Theorem 1.7. Let G = 1 + A + B + C + D, H = + A B C D C CD D. The cyclic inequality f 4 (x, y, z) 0 holds for all real numbers x, y, z if the following two conditions are satisfied: (a) 1 + A + B + C + D 0; (b) there exists a real number t ( 3, 3) such that f(t) 0, where f(t) = Gt 3 (6 + A + B + 3C + 3D)t + (1 + C + D)Gt + H. In this paper, we will establish some very strong sufficient conditions that the inequality holds for all nonnegative real numbers x, y, z. f 4 (x, y, z) 0. Main Results The main result of this paper is given by the following two theorems. Theorem.1. The inequality f 4 (x, y, z) 0 holds for all nonnegative real numbers x, y, z if 1 + A + B + C + D 0 one of the following two conditions is fulfilled: (a) 3(1 + A) C + CD + D ; (b) 3(1 + A) < C + CD + D, 5 + A + C + D 0, f 4 (x, 1, 0) 0, h 3 (x) 0 for all x 0, where h 3 (x) = (4 + C + D)(x 3 + 1) + (A + C D 1)x + (A C + D 1)x. Theorem.. The inequality f 4 (x, y, z) 0 holds for all nonnegative real numbers x, y, z if 1 + A + B + C + D 0 one of the following two conditions is fulfilled: (a) 3(1 + A) C + CD + D ; (b) 3(1 + A) < C + CD + D, there is t 0 such that (C + D)t + 6t + C + D (t 4 + t + 1)(C + CD + D 3 3A).

4 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), Remark.3. If the sufficient conditions in Theorem.1 or Theorem. are fulfilled, then the following sharper inequality holds for all x, y, z 0: f 4 (x, y, z) (1 + A + B + C + D)xyz x. This claim is true because B 1 A C D, on the other h, Theorems.1. remain valid by replacing B with 1 A C D. Therefore, if 1 + A + B + C + D > 0 the other sufficient conditions in Theorem.1 or Theorem. are fulfilled, then the inequality f 4 (x, y, z) 0 becomes an equality only when one of x, y, z is zero; that is, for x = βy z = 0 (or any cyclic permutation), where β is a double positive root of the polynomial f 4 (x, 1, 0) (see the proof of Theorem.1) or h 4 (t) = [(C + D)t + 6t + C + D] 4(t 4 + t + 1)(C + CD + D 3 3A) (see the proof of Theorem.). Remark.4. Consider the main case when 1 + A + B + C + D = 0. In the case (a) of Theorem.1 Theorem., the inequality f 4 (x, y, z) 0 holds for all real numbers x, y, z, the equality conditions (including the case x = y = z) are given by Proposition 1.5 Proposition 1.6. In the case (b) of Theorem.1 Theorem., the inequality f 4 (x, y, z) 0 holds for all nonnegative real numbers x, y, z, but does not hold for all real numbers x, y, z. Equality holds for x = y = z, for x = βy z = 0 (or any cyclic permutation), where β is a double positive root of the polynomial f 4 (x, 1, 0) (see the proof of Theorem.1) or h 4 (t) (see the proof of Theorem.). Conjecture.5. If 1 + A + B + C + D = 0, then the conditions in Theorem.1 Theorem. are necessary sufficient to have f 4 (x, y, z) 0 for all x, y, z 0. Conjecture.6. If the inequality f 4 (x, y, z) 0 does not hold for all real numbers x, y, z, then the conditions in Theorem.1 Theorem. are necessary sufficient to have f 4 (x, y, z) 0 for all x, y, z Proof of Theorem.1 Since Let us define f 4 (x, y, z) = x 4 + A x y (1 + A + B + C + D)xyz x + C x 3 y + D xy 3. f 4 (x, y, z) f 4 (x, y, z) for all x, y, z 0, it suffices to prove that f 4 (x, y, z) 0. Assume that x = min{x, y, z}, use the substitution y = x + p, z = x + q, where p, q 0. From x 4 = 3x 4 + 4(p + q)x 3 + 6(p + q )x + 4(p 3 + q 3 )x + p 4 + q 4, x y = 3x 4 + 4(p + q)x 3 + (p + q) x + pq(p + q)x + p q, xyz x = 3x 4 + 4(p + q)x 3 + (p + 5pq + q )x + pq(p + q)x, x 3 y = 3x 4 + 4(p + q)x 3 + 3(p + pq + q )x + (p 3 + 3p q + q 3 )x + p 3 q,

5 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), xy 3 = 3x 4 + 4(p + q)x 3 + 3(p + pq + q )x + (p 3 + 3pq + q 3 )x + pq 3, we get where f 4 (x, y, z) = A 1 (p, q)x + B 1 (p, q)x + C 1 (p, q) := h(x), A 1 (p, q) = (5 + A + C + D)(p pq + q ), B 1 (p, q) = (4 + C + D)(p 3 + q 3 ) + (A + C D 1)p q + (A C + D 1)pq, C 1 (p, q) = p 4 + Cp 3 q + Ap q + Dpq 3 + q 4. As we have shown in [3], the inequality h(x) 0 holds for all real x all p, q 0 if 3(1+A) C +CD+D. Assume now that 3(1 + A) < C + CD + D. Clearly, the inequality h(x) 0 holds for all nonnegative real x if A 1 (p, q) 0, B 1 (p, q) 0 C 1 (p, q) 0 for all p, q 0. Clearly, these inequality are respectively equivalent to 5 + A + C + D 0, h 3 (x) 0 for all x 0 f 4 (x, 1, 0) 0 for all x Proof of Theorem. (a) By Corollary 1.4, if 3(1 + A) C + CD + D B = 1 A C D, then f 4 (x, y, z) 0 for all real numbers x, y, z, so the more for all nonnegative real numbers x, y, z. Since the polynomial f 4 is increasing in B, the inequality f 4 (x, y, z) 0 holds also for all B 1 A C D. (b) The main idea is to find a sharper cyclic homogeneous inequality of degree four x 4 + A 1 x y + B 1 xyz x + C 1 x 3 y + D 1 xy 3 0, such that Let us define where with 1 + A 1 + B 1 + C 1 + D 1 = 0. f 4 (x, y, z) = f 4 (x, y, z) g(x, y, z), g(x, y, z) = yz(px + qy qtz) + zx(py + qz qtx) + xy(pz + qx qty), t 0, q = 4 C + CD + D 3 3A t 4 + t + 1 > 0, p = q(t 1) A + B + C + D. Since g(x, y, z) 0, it suffices to prove that f 4 (x, y, z) 0. We can write f 4 (x, y, z) in the form f 4 (x, y, z) = x 4 + A 1 x y + B 1 xyz x + C 1 x 3 y + D 1 xy 3, where A 1 = A + q t, B 1 = B p(p + q qt), C 1 = C q, D 1 = D q t. Since 1 + A 1 + B 1 + C 1 + D 1 = 1 + A + B + C + D (p + q qt) = 0,

6 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), according to Corollary 1.4, it suffices to show that 3(1 + A 1 ) C1 + C 1D 1 + D1. Write this inequality as (C + D)t + 6t + C + D q (t 4 + t + 1) + 1 q (C + CD + D 3 3A), (C + D)t + 6t + C + D (t 4 + t + 1)(C + CD + D 3 3A). By the hypothesis in (b), there is t 0 such that the last inequality is true. Thus, the proof is completed. 5. Applications Application 5.1. Let x, y, z be nonnegative real numbers. If k 0, then ([] [7]) x 4 + (k ) x y + (1 k )xyz x k( x 3 y xy 3 ). Proof. Write the inequality as f 4 (x, y, z) 0, where A = k, B = 1 k, C = k, D = k, 1 + A + B + C + D = 0. First Solution. We will show that the condition (b) in Theorem.1 is fulfilled. Since C + CD + D 3(1 + A) = k + 3 > A + C + D = k + 3 > 0, we only need to show that f 4 (x, 1, 0) 0 h 3 (x) 0 for all x 0. We have f 4 (x, 1, 0) = x 4 kx 3 + (k )x + kx + 1 = (x kx 1) 0, For 0 x < 1, we get h 3 (x) = 4(x 3 + 1) + (k 6k 3)x + (k + 6k 3)x. h 3 (x) = 4(x 3 + 1) + (k 3)x(1 + x) + 6kx(1 x) 4(x 3 + 1) + (k 3)x(1 + x) Also, for x 1, we get 4(x 3 + 1) 4x(1 + x) = 4(x + 1)(x 1) > 0. h 3 (x) = 4(x 1) 3 + (k 3) x + (k + 6k 15)x + 8 = 4(x 1) 3 + (k 3) (x 1) + 3(k 1) x k + 6k 1 = 4(x 1) 3 + (k 3) (x 1) + 3(k 1) (x 1) + (k + 1) > 0. The polynomial f 4 (x, 1, 0) has the double positive real root β = k + k + 4. Therefore, according to Remark.4, equality holds for x = y = z, also for x = 0 y z = k + k + 4 (or any cyclic permutation). Second Solution. We will show that the condition (b) in Theorem. is fulfilled. Since C + CD + D 3(1 + A) = k + 3 > 0, we only need to show that there exists t 0 such that kt + 3t k (k + 3)(t 4 + t + 1).

7 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), This is true if h 4 (t) 0, where kt + 3t k 0 Clearly, for we have h 4 (t) = 0 h 4 (t) = (kt + 3t k) (k + 3)(t 4 + t + 1) = (t kt 1). t = k + k + 4, kt + 3t k = k(kt + 1) + 3t k = (k + 3)t > 0. Since the polynomial h 4 (t) has the double positive real root β = k + k + 4, according to Remark.4, equality holds for x = y = z, also for x = 0 y z = k + k + 4 (or any cyclic permutation). Remark. For k = 1, we get the inequality x 4 + y 4 + z 4 x y y z z x (x 3 y + y 3 z + z 3 x xy 3 yz 3 zx 3 ), with equality for x = y = z, for x = 0 y z = Also, for k =, we get the inequality (or any cyclic permutation). x 4 + y 4 + z 4 xyz(x + y + z) (x 3 y + y 3 z + z 3 x xy 3 yz 3 zx 3 ), with equality for x = y = z, for x = 0 y z = + 6 (or any cyclic permutation). Application 5.. If x, y, z are nonnegative real numbers, then ([]) x 4 + y 4 + z 4 + 5(x 3 y + y 3 z + z 3 x) 6(x y + y z + z x ). Proof. Write the inequality as f 4 (x, y, z) 0, where A = 6, B = 0, C = 5, D = 0, 1 + A + B + C + D = 0. First Solution. We will show that the condition (b) in Theorem.1 is fulfilled. Since C + CD + D 3(1 + A) = A + C + D = 9, we only need to show that f 4 (x, 1, 0) 0 h 3 (x) 0 for all x 0. We have f 4 (x, 1, 0) = x 4 + 5x 3 6x + 1 = (x 1) 4 + x(3x ) > 0 For 0 x < 1, we get for x 1, we get h 3 (x) = 3(3x 3 + x 4x + 3). 3x 3 + x 4x + 3 (x 1)(x 3) > 0, 3x 3 + x 4x + 3 4x(x 1) + 3 > 0.

8 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), Since the polynomial f 4 (x, 1, 0) has no double positive real root, equality holds only for x = y = z (see Remark.4). Second Solution. We will show that the condition (b) in Theorem. is fulfilled. Since we only need to show that there is t 0 such that Indeed, for t = 3/, we get C + CD + D 3(1 + A) = 40, 10t + 6t (t 4 + t + 1). 10t + 6t (t 4 + t + 1) = = According to Remark.4, equality holds for x = y = z. Application 5.3. If x, y, z are nonnegative real numbers, then 9 ( > 0. 3(x 4 + y 4 + z 4 ) + 4(xy 3 + yz 3 + zx 3 ) 7(x 3 y + y 3 z + z 3 x). Proof. Write the inequality as f 4 (x, y, z) 0, where A = 0, B = 0, C = 7 3, D = 4 3, 1 + A + B + C + D = 0. First Solution. We will prove that the condition (b) in Theorem.1 is fulfilled. Since C + CD + D 3(1 + A) = A + C + D =, we only need to show that f 4 (x, 1, 0) 0 h 3 (x) 0 for all x 0. We have ( f 4 (x, 1, 0) = x(x + 1)(3x 5) + 5 x 13 ) > 0, h 3 (x) = 3x 3 7x + 4x + 3. For 0 x 1 x 4, we get 3 for 1 x 3, we get 3x 3 7x + 4x + 3 > 3x 3 7x + 4x = x(x 1)(3x 4) 0, 3x 3 7x + 4x + 3 4x + 4x + 3 = (x + 1)(3 x) 0. Since the polynomial f 4 (x, 1, 0) has no double positive real root, equality holds only for x = y = z (see Remark.4). Second Solution. We will prove that the condition (b) in Theorem. is fulfilled. Since C + CD + D 3(1 + A) = 10 9

9 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), we only need to show that there exists t 0 such that Indeed, for t =, we get t + 18t 10 10(t 4 + t + 1). t + 18t 10 10(t 4 + t + 1) = = According to Remark.4, equality holds for x = y = z > 0. Application 5.4. If x, y, z are nonnegative real numbers, then ([1]) ( ) 4 x 4 + y 4 + z xyz(x + y + z) 4 7 (x 3 y + y 3 z + z 3 x). 7 Proof. Write the inequality as f 4 (x, y, z) 0, where A = 0, B = 4 7 1, C = 4 7, D = 0, 1 + A + B + C + D = 0. First Solution. We will show that the condition (b) in Theorem.1 is fulfilled. Since C + CD + D 3(1 + A) = > 0, 5 + A + C + D = > 0, we only need to show that f 4 (x, 1, 0) 0 h 3 (x) 0 for all x 0. We have f 4 (x, 1, 0) = x x = (x 3) ( x + 7 x ) 0 h 3 (x) = 4x 3 x x (x 3 + x x + 1). Since we get x 3 + x x + 1 x x + 1 > < 9 5, 5h 3 (x) > 5(4x 3 x x + 4) 9(x 3 + x x + 1) = 11x 3 3x + 4x + 11 ( = 11x x 3 ) + 10x 83 4 x x 83 4 x + 11 ( = 10 x 83 ) > 0. The polynomial f 4 (x, 1, 0) has the double positive real root β = 3. Therefore, according to Remark.4, equality holds for x = y = z, also for x = 0 y z = 3 (or any cyclic permutation).

10 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), Second Solution. We will show that the condition (b) in Theorem. is fulfilled. Since C + CD + D 3(1 + A) = > 0, we only need to show that there exists t 0 such that t + 3 7t 4 (16 9 3)(t 4 + t + 1). This is true if h 4 (t) 0, where Since we have h 4 (t) = 0 for t = 3, when t + 3 7t 4 0 h 4 (t) = ( t + 3 7t 4) (16 9 3)(t 4 + t + 1). h 4 (t) = 3(t 3) [(3 3 4)t 3(4 3)t + 3], t + 3 7t 4 = 5 3 > 0. The polynomial h 4 (t) has the double positive real root β = 3. Therefore, according to Remark.4, equality holds for x = y = z, also for x = 0 y z = 3 (or any cyclic permutation). Application 5.5. If x, y, z are nonnegative real numbers, then ([7]) x 4 + y 4 + z (x 3 y + y 3 z + z 3 x) 47 4 (x y + y z + z x ). Proof. Write the inequality as f 4 (x, y, z) 0, where A = 47 17, B = 0, C = 15, D = 0, 1 + A + B + C + D = 4 4. First Solution. We will show that the condition (b) in Theorem.1 is fulfilled. Since C + CD + D 3(1 + A) = 109 4, 5 + A + C + D = 93 4, we only need to show that f 4 (x, 1, 0) 0 h 3 (x) 0 for all x 0. We have f 4 (x, 1, 0) = x x x + 1 = 1 4 (x 1) (x + 16x + 4) 0. h 3 (x) = 19(x 3 + 1) x x > x 8x = 14(x 1) 0. According to Remark.3, since the polynomial f 4 (x, 1, 0) has the double nonnegative real root β = 1, equality holds for x = 0 y = z (or any cyclic permutation).

11 Y. Zhou, V. Cirtoaje, J. Nonlinear Sci. Appl. 6 (013), Second Solution. We will show that the condition (b) in Theorem. is fulfilled. Since C + CD + D 3(1 + A) = 109 4, we only need to show that there is t 0 such that This is true if h 4 (t) 0, where Since we have h 4 (t) = 0 for t = 1. 15t + 6t (t 4 + t + 1). h 4 (t) = (15t + 6t + 30) 109(t 4 + t + 1). h 4 (t) = 3(t 1) (67t + 5t + 43), According to Remark.3, since the polynomial h 4 (t) has the double nonnegative real root β = 1, equality holds for x = 0 y = z (or any cyclic permutation). Application 5.6. If x, y, z are nonnegative real numbers such that x + y + z = 5 (xy + yz + zx), then x 4 + y 4 + z (x3 y + y 3 z + z 3 x). Proof. We see that equality holds for x = 0, y =, z = 1 (or any cyclic permutation). Since it suffices to show that In addition, since it suffices to show that which is equivalent to x 4 + y 4 + z 4 (x + y + z ) (xy + yz + zx) = 17 4 (xy + yz + zx), (xy + yz + zx) x 3 y + y 3 z + z 3 x. 36(xy + yz + zx) = [6(xy + yz + zx) = [(x + y + z ) + xy + yz + zx], It suffices to show that f 4 (x, y, z) 0, where with Since [(x + y + z ) + xy + yz + zx] 18(x 3 y + y 3 z + z 3 x), 4 x x y + 6xyz x + 4 xy 3 14 x 3 y. f 4 (x, y, z) = 4 x x y 3xyz x 14 x 3 y + 4 xy 3. A = 9 4, B = 3 4, C = 7, D = 1, 1 + A + B + C + D = (1 + A) C CD D = 0, the condition (a) in Theorem.1 Theorem. is fulfilled.

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