On a class of three-variable inequalities. Vo Quoc Ba Can

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1 On a class of three-variable inequalities Vo Quoc Ba Can 1 Theem Let a, b, c be real numbers satisfying a + b + c = 1 By the AM - GM inequality, we have ab + bc + ca 1, therefe setting ab + bc + ca = 1 q (q 0), we will find the maximum and minimum values of abc in terms of q If q = 0, then a = b = c = 1 1, therefe abc = If q 0, then (a b) + (b c) + (c a) > 0 Consider the function f(x) = (x a)(x b)(x c) = x x + 1 q x abc f (x) = x x + 1 q whose zeros are x 1 = 1 + q, and x = 1 q We can see that f (x) < 0 f x < x < x 1 and f (x) > 0 f x < x x > x 1 Furtherme, f(x) has three zeros: a, b, and c Then ( ) 1 q f = (1 q) (1 + q) abc 0 and Hence and we obtain ( ) 1 + q f (1 + q) (1 q) = (1 + q) (1 q) abc 0 abc (1 q) (1 + q) Theem 11 If a, b, c are arbitrary real numbers such that a + b + c = 1, then setting ab + bc + ca = 1 q (q 0), the following inequality holds Or, me general, (1 + q) (1 q) abc (1 q) (1 + q) Theem 1 If a, b, c are arbitrary real numbers such that a + b + c = p, then setting ab + bc + ca = p q (q 0) and r = abc, we have (p + q) (p q) r (p q) (p + q) This is a powerful tool since the equality holds if and only if (a b)(b c)(c a) = 0 Mathematical Reflections (007) 1

2 Here are some identities which we can use with this theem Applications a + b + c = p + q a + b + c = pq + r ab(a + b) + bc(b + c) + ca(c + a) = p(p q ) r (a + b)(b + c)(c + a) = p(p q ) r a b + b c + c a = (p q ) pr ab(a + b ) + bc(b + c ) + ca(c + a ) = (p + q )(p q ) pr a 4 + b 4 + c 4 = p4 + 8p q + q 4 + 4pr 1 Let a, b, c be positive real numbers such that a + b + c = 1 Prove that 1 a + 1 b (ab + bc + ca) 5 c Solution We can easily check that q [0, 1], by using the theem we have LHS = 1 q r + 16(1 q ) (1 + q) (1 q)(1 + q) + 16(1 q ) = q (4q 1) (1 q)(1 + q) The inequality is proved Equality holds if and only if a = b = c = 1 a = 1, b = c = 1 4 permutations and their [Vietnam 00] Let a, b, c be real numbers such that a + b + c = Prove that (a + b + c) abc 10 Solution The condition can be rewritten as p + q = Using our theem, we have We need to prove that This follows from, equivalently, LHS = p r p (p + q) (p q) p(5q + ) 0 q (0 q ) p (5q + ), = p(5q + ) + q (q ) (q 4 + 1q + 4q + 146q + 1) 0 The inequality is proved Equality holds if and only if a = b =, c = 1 and their permutations [Vo Quoc Ba Can] F all positive real numbers a, b, c, we have a + b + b + c + c + a ab + bc + ca + 11 c a b a + b + c 17 Mathematical Reflections (007)

3 Solution Because the inequality is homogeneous, without loss of generality, we may assume that p = 1 Then q [0, 1] and the inequality can be rewritten as 1 q 1 q + 11 r 1 + q 0 Using our theem, it suffices to prove 1 q q 0 (1 + q) (1 q)(1 + q) = 40q (1 q)(1 + q) If 40q + 11q , q , it is trivial If q <, we have 11(1 q ) (1 + q ) ( 40q + 11q + 11) (1 q) (1 + q) = q (11 110q + 55q + 748q 18q 4 ) (1 + q )(1 q) (1 + q) On the other hand, ( q + 55q + 748q 18q 4 = q 4 q q + 55 q ) q 18 q 4 ( 11 (/) (/) + 55 (/) / 18 The inequality is proved Equality occurs if and only if a = b = c 4 [Vietnam TST 16] Prove that f any a, b, c R, the following inequality holds (a + b) 4 + (b + c) 4 + (c + a) (a4 + b 4 + c 4 ) ) = q4 0 Solution If p = 0 the inequality is trivial, so we will consider the case p 0 Without loss of generality, we may assume p = 1 The inequality becomes Using our theem, we have q 4 + 4q r 0 q 4 + 4q r q 4 + 4q (1 q) (1 + q) = q (q 4) + q The inequality is proved Equality holds only f a = b = c = 0 5 [Pham Huu Duc, MR1/007] Prove that f any positive real numbers a, b, and c, b + c c + a a + b a + b + c a b c abc Solution By Holder s inequality, we have ( ) ( b + c a ) ( 1 a (b + c) ) 1 a It suffices to prove that ( ) 1 6(a + b + c) 1 a abc a (b + c) Mathematical Reflections (007)

4 Setting x = 1 a, y = 1 b, z = 1 c, the inequality becomes (x + y + z) 6 xyz(xy + yz + zx) x y + z, (x + y + z) 6 xyz(xy + yz + zx)((x + y + z) (x + y + z)(xy + yz + zx) + xyz) (x + y)(y + z)(z + x) By the AM - GM inequality, (x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) xyz 8 (x + y + z)(xy + yz + zx) It remains to prove that 4(x + y + z) 4 xyz((x + y + z) (x + y + z)(xy + yz + zx) + xyz) Setting p = x + y + z, xy + yz + zx = p q Applying our theem, it suffices to prove that 4p 4 (p q) (p + q) (p q 0), the inequality becomes 4p 4 xyz(p + pq + xyz) ( p + pq + (p ) q) (p + q), 4p 4 (p q) (p + q)(p + 6pq + (p q) (p + q)) Setting u = p q p+q 1, the inequality is equivalent to 4(u + 1) 4 u (4u + 5u 6 + u + 1), f(u) = (u + 1) 4 u (4u + 5u 6 + u + 1) 4 f (u) = (u + 1) (u 1)(u 1)(u 6 + u 1) u (u + 1) (4u 6 + u + 1) f (u) = 0 u = 1, u = 1, u = 1 Now, we can easily verify that f(u) min f 1, f(1) = 4, which is true The inequality is proved Equality holds if and only if a = b = c 6 [Darij Grinberg] If a, b, c 0, then a + b + c + abc + 1 (ab + bc + ca) Mathematical Reflections (007) 4

5 Solution Rewrite the inequality as 6r + + 4q p 0 If q p, it is trivial If p q, using the theem, it suffices to prove that (p q)(p + q) + + 4q p 0, (p ) (p + ) q (q + p 18) If p, we have q + p 4p 18, therefe the inequality is true If p, we have q (q + p 18) 4q (p ) p (p ) = (p ) (p + ) < (p ) (p + ) The inequality is proved Equality holds if and only if a = b = c = 1 7 [Schur s inequality] F any nonnegative real numbers a, b, c, a + b + c + abc ab(a + b) + bc(b + c) + ca(c + a) Solution Because the inequality is homogeneous, we can assume that a + b + c = 1 Then q [0, 1] and the inequality is equivalent to r + 4q 1 0 If q 1, it is trivial If q 1, by the theem we need to prove that (1 + q) (1 q) + 4q 1 0, q (1 q) 0, which is true Equality holds if and only if a = b = c a = b, c = 0 and their permutations 8 [Pham Kim Hung] Find the greatest constant k such that the following inequality holds f any positive real numbers a, b, c a + b + c k(ab + bc + ca) + (a + b)(b + c)(c + a) (a + b + c) 8 + k Solution F a = b = 1 + and c = 1, we obtain k (+ ) 8 = k 0 We will prove that this is the desired value Let k 0 be a constant satisfying the given inequality Without loss of generality, assume that p = 1 Then q [0, 1] and the inequality becomes (r + q ) r + 1 q + k 0(1 q ) 8 + k 0 It is not difficult to verify that this is an increasing function in terms of r If q 1, we have V T q 1 q + k 0(1 q ) 1 + k k 0 (since this is an increasing function in terms of q 1 4 ) If q 1, using our theem, it suffices to prove that ((1 + q) (1 q) + q ) (1 + q) (1 q) + (1 q ) + k 0(1 q ) 8 + k 0 LHS RHS = q ( + ) ( 1 q ) ( q + ) 8(q + 1)(q ) 0 The inequality is proved, and we conclude that k max = k 0 Mathematical Reflections (007) 5

6 [Pham Huu Duc] F all positive real numbers a, b and c, ( ) 1 a + bc + 1 b + ca + 1 (a + b + c) 1 c + ab (ab + bc + ca) a + b + 1 b + c + 1 c + a Solution Because the inequality is homogeneous, we may assume that p = } 1 Then q [0, 1] and by the AM - GM and Schur s inequalities, we have (1 q ) r max {0, 1 4q After expanding, we can rewrite the given inequality as f(r) = 486( q )r + (q q 4 5q + 4)r + (4q 1)(11q 4 4q + )r + q (1 q ) (q 4 + 8q 1) 0 f (r) = ( 16( q )r + 6(q q 4 5q + 4)r + (4q 1)(11q 4 4q + )) f (r) = 54( 54( q )r + q q 4 5q + 4) 54( (1 q ) ( q ) + q q 4 5q + 4) = 16(q q 4 + q + ) > 0 Hence f (r) is an increasing function Now, if 1 q, then f (r) f (0) = (4q 1)(11q 4 4q + ) 0 If 1 q, then ( ) 1 4q f (r) f = (1 4q )(q + )(q q + 6) 0 In any case, f(r) is an increasing function If 1 q, then f(r) f(0) = q (1 q ) (q 4 + 8q 1) 0, and we are done If 1 q, using our theem, we have ( (1 + q) ) (1 q) f(r) f The proof is complete Equality holds if and only if a = b = c = 1 81 q ( q)(q + 1) (6q + 4q 7q + 4)(5q q + ) 0 10 [Nguyen Anh Tuan] Let x, y, z be positive real numbers such that xy + yz + zx + xyz = 4 Prove that x + y + z xy + yz + zx ((x y) + (y z) + (z x) ) Solution Since x, y, z > 0 and xy + yz + zx + xyz = 4, there exist a, b, c > 0 such that x = a b+c, y = b c+a, z = c a+b The inequality becomes P (a, b, c) = (a + b + c) (a b ) (a + b) (b + c) (c + a) 6 a(a + b)(a + c) ab(a + b) Because the inequality is homogeneous we can assume that p = 1 computations, we can rewrite the inequality as Then q [0, 1] and after some f(r) = 7r + (q 1)r + (6q 4 4q + 1)r + (q 1)(1q 4 5q + 1) 0 f (r) = (r(81r + 44q ) + 6q 4 4q + 1) Mathematical Reflections (007) 6

7 By Schur s inequality, 81r + 44q (1 4q ) + 44q = 1 + q > 0 Hence f (r) 0, and f(r) is an increasing function Then by our theem we have ( (1 q) ) (1 + q) f(r) f = q (q 1)(q + ) (4q q + 15q 7q + 1) 0 The inequality is proved Equality holds if and only if x = y = z 11 [Nguyen Anh Tuan] F all nonnegative real numbers a, b, c (a ab + b )(b bc + c )+ (b bc + c )(c ca + a )+ (c ca + a )(a ab + b ) a +b +c Solution After squaring both sides, we can rewrite the inequality as ( ) ( ) ( ) (a ab + b ) a ab + b ab a a b By the AM - GM inequality, a ab + b 1 (a + b), b bc + c 1 (b + c), c ca + a 1 (c + a) It suffices to prove that ( ) ( ) ( ) (a ab + b ) a ab a a b Because this inequality is homogeneous, we can assume p = 1 Then q [0, 1] and the inequality is equivalent to 7r + (1 10q )r + q (1 q ) 6r + q (1 q ), f(r) = 4r 1r(q 4 11q + 1) q (4 q )(1 q ) 0 It is not difficult to verify that f(r) is a convex function, then using our theem, we have { ( (1 q) )} (1 + q) f(r) max f(0), f Furtherme, f(0) = q (4 q )(1 q ) 0, ( (1 q) ) (1 + q) f = 1 q (q 1) (q + )(q + q + ) 0 Our proof is complete Equality holds if and only if a = b = c a = t 0, b = c = 0, and their permutations Vo Quoc Ba Can: Can Tho University, Vietnam Mathematical Reflections (007) 7

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