Two Applications of RCF, LCF, and EV Theorems Vasile Cirtoaje Abstract In this paper we present two new and difficult symmetric inequalities with right convex and left concave functions, as applications of RCF-Theorem and LCF-Theorem from [1], [2] and [3] Moreover, we show that both inequalities can be also proved using the Equal Variable Theorem from [2] and [5] Proposition If a1 , a2 , , an , n ≤ 81 are nonegative real numbers such that a61 + a62 + · · · + a6n = n, then a21 + a22 + · · · + a2n ≤ a51 + a52 + · · · + a5n Proof By letting an = 1, we obtain the initial statement but for n − numbers Thus it suffices to prove the inequality for n = 81 Let us make the following substitution: xi = ai6 for all i Now we have to prove that 1 5 x13 + x23 + · · · + xn3 ≤ x16 + x26 + · · · + xn6 when x1 + x2 + · · · + x81 = 81 This inequality is equivalent to f (x1 ) + f (x2 ) + · · · + f (x81 ) ≤ 81 · f x1 + x2 + · · · + x81 81 , where f (u) = u − u , u ≥ The second derivative of f (u) is f (u) = 1 −5 u 5u − 36 x1 + x2 + · · · + x81 = 81 Thus by the LCF theorem, it suffices to prove the inequality for It follows that f is concave for u ≤ s, where s = x1 = x2 = · · · = x80 ≤ ≤ x81 This requires to prove the original inequality for a1 = a2 = · · · = a80 ≤ ≤ a81 Let us rewrite the original inequality in the homogeneous form 81 a51 + a52 + · · · + a581 Mathematical Reflections (2008) ≥ a61 + a62 + · · · + a681 a21 + a22 + · · · + a281 Since the case a1 = a2 = · · · = a80 = is trivial, we assume that a1 = a2 = · · · = a80 = Let a81 = x, then the inequality becomes 81(80 + x5 )2 ≥ (80 + x6 )(80 + x2 )2 which is equivalent to (x − 1)2 (x − 2)2 (x6 + 6x5 + 21x4 + 60x3 + 75x2 + 60x + 20) ≥ 0, and clearly is true For a1 ≤ a2 ≤ · · · ≤ a81 , equality occurs in the above homogeneous inequality when a1 = a2 = · · · = a81 or a1 = a2 = · · · = 12 a81 In the original inequality, equality occurs when a1 = a2 = · · · = an = Moreover, for n = 81, √ equality occurs when a1 = a2 = · · · = a80 = 34 and a81 = Remark The inequality is not valid for n > 81 To prove this is is enough to let a1 = a2 = · · · = an−1 = and an = in the homogeneous inequality n a51 + a52 + · · · + a5n ≥ a61 + a62 + · · · + a6n a21 + a22 + · · · + a2n We would get that (n − 1)(81 − n) ≥ 0, clearly false for n > 81 Remark We can also prove the inequality in Proposition using the Equal Variable Theorem ([2], [5]) According to the EV theorem, the following statement holds: 1 If ≤ x1 ≤ x2 ≤ ≤ x81 such that x1 + x2 + · · · + x81 = 81 and x13 + x23 + · · · + x81 5 is constant, then the sum x16 + x26 + · · · + x81 is minimal whenever x1 = x2 = · · · = x80 ≤ x81 Proposition Let a1 , a2 , , an be positive real numbers such that a1 a2 · · · an = If m ≥ n − 1, then m m am + a2 + · · · + an + n(2m − n) ≥ (2m − n + 1) 1 + + ··· + a1 a2 an Proof Since the case n = and m = is trivial, we may assume that m > Let = exi for all i We have to prove that emx1 + emx2 + · · · + emxn + n(2m − n) ≥ (2m − n + 1)(e−x1 + e−x2 + · · · + e−xn ) for x1 + x2 + · · · + xn = This inequality is equivalent to f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ n · f Mathematical Reflections (2008) x1 + x2 + · · · + xn n , where f (u) = emu + 2m − n − (2m − n + 1)e−u , u ∈ R n We will prove that f (u) is right convex for u ≥ s, where s = x1 +x2 +···+x = 0; or n f (u) ≥ for u ≥ Taking the second derivative of f (u), we get f (u) = e−u [m2 e(m+1)u − 2m + n − 1] > 0, because m2 e(m+1)u − 2m + n − ≥ m2 − 2m + n − = (m − 1)2 + n − > According to the RCF theorem, it suffices to prove the inequality above for x2 = x3 = · · · = xn ≥ or, equivalently, the original inequality for a2 = · · · = an ≥ : xm + (n − 1)y m + n(2m − n) ≥ (2m − n + 1) n−1 + x y for m ≥ n − 1, < x ≤ ≤ y and xy n−1 = Let us rewrite the inequality as f (y) ≥ 0, where f (y) = y m(n−1) We have f (y) = + (n − 1)y m + n(2m − n) − (2m − n + 1) y n−1 + (n−1)g(y) , y mn−m+1 n−1 y g(y) = m(y mn−1 ) − (2m − n + 1)y mn−m−1(y n −1) , and g (y) = y mn−m−2 h(y), where h(y) = m2 ny m+1 − (2m − n + 1)[(m + 1)(n − 1)y n − mn + m + 1] h (y) = (m + 1)ny n−1 [m2 y m−n+1 − (2m − n + 1)(n − 1)] If m = n − and n ≥ 3, then h(y) = n(n − 1)(n − 2) > Otherwise, if m > n − and n ≥ 2, then m2 y m−n+1 − (2m − n + 1)(n − 1) ≥ m2 − (2m − n + 1)(n − 1) = (m − n + 1)2 > 0, and hence h (y) > for y ≥ Therefore, h(y) is strictly increasing on [1, ∞), and h(y) ≥ h(1) = n[(m − 1)2 + n − 2] > for y ≥ Since h(y) > implies g y > 0, it follows that g(y) is strictly increasing on [1, ∞) Then g(y) ≥ g(1) for y ≥ 1, and from y mn−m+1 f (y) = (n − 1)g(y) ≥ it follows that f (y) is strictly increasing on [1, ∞) Consequently, f (y) ≥ f (1) = for y ≥ For n = and m = 1, the original inequality becomes an equality Otherwise, equlity occurs if and only if a1 = a2 = · · · = an = Mathematical Reflections (2008) Remark For m = n − 1, the following statement is true: If a1 , a2 , , an are positive real numbers such that a1 a2 · · · an = 1, then an−1 + an−1 + · · · + an−1 + n(n − 2) ≥ (n − 1) n 1 + + ··· + a1 a2 an This inequality follows from the Generalized Popoviciu’s Inequality If f is a convex function on an interval I and a1 , a2 , , an ∈ I, then f (a1 ) + · · · + f (an ) + n(n − 2)f where bj = n j=i aj , n−1 a1 + · · · + an n ≥ (n − 1) (f (b1 ) + · · · + f (bn )) , for all i Consider the convex function f (x) = ex , replace a1 , a2 , , an with (n−1) ln a1 , (n− 1) ln a2 , , (n − 1) ln an , and you get the desired result (see [4]) Remark Replacing a1 , a2 , , an by x11 , x12 , , x1n the inequality in Proposition becomes 1 + m + · · · + m + (2m − n)n ≥ (2m − n + 1)(x1 + x + + · · · + xn ), xm x x n where x1 x2 · · · xn = We can also prove the inequality by the Equal Variable theorem: If < x1 ≤ x2 ≤ · · · ≤ xn such that x1 + x2 + · · · + xn = constant and x1 x2 · · · xn = 1, then the sum xm + xm +···+ xm n is minimal when < x1 = x2 = · · · = xn−1 ≤ xn References [1] V Cirtoaje, A generalization of Jensens Inequality, Gazeta Matematica Seria A, (2005), 124-138 [2] V Cirtoaje, Algebraic Inequalities Old and New Method, Gil Publishing House, 2006 [3] V Cirtoaje, Four Applications of RCF and LCF Theorems, Mathematical Reflections, (2007).[http://reflections.awesomemath.org/2007 1.html] [4] V Cirtoaje, Two Generalizations of Popovicius Inequality, Crux Math., (2005), 313-318 [5] V Cirtoaje, The Equal Variable Method, Journal of Inequalities in Pure and Applied Mathematics, (1) (2007), Art 15 [http://jipam.vu.edu.au/article.php?sid=828] Vasile Cartoaje Petroleum-Gas University, Ploiesti, Romania email: vcirtoaje@upg-ploiesti.ro Mathematical Reflections (2008)