Two Applications of RCF, LCF, and EV Theorems

Two Applications of RCF, LCF, and EV TheoremsVasile Cirtoaje Abstract In this paper we present two new and difficult symmetric inequalities with right convex and left concave functions,

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 1 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 − 1 numbers Thus it suffices to prove the inequality for n = 81 Let us make the following substitution: xi= a

1 6

i for all i Now we have to prove that x

1 3

1 + x

1 3

2 + · · · + x

1 3

n ≤ x

5 6

1 + x

5 6

2 + · · · + x

5 6

n

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) = u13 − u56, u ≥ 0 The second derivative of f (u) is

f00(u) = 1

36u

− 5 5u1 − 8

It follows that f is concave for u ≤ s, where s = x1+ x2+ · · · + x81

Thus by the LCF theorem, it suffices to prove the inequality for

x1= x2 = · · · = x80≤ 1 ≤ x81 This requires to prove the original inequality for a1 = a2= · · · = a80≤ 1 ≤ a81 Let

us rewrite the original inequality in the homogeneous form

81 a51+ a52+ · · · + a581
2≥ a61+ a62+ · · · + a681
a21+ a22+ · · · + a281
2

Since the case a1 = a2 = · · · = a80 = 0 is trivial, we assume that a1 = a2 = · · · =

a80= 1 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 homoge-neous inequality when a1 = a2 = · · · = a81 or a1 = a2 = · · · = 12a81 In the original inequality, equality occurs when a1 = a2 = · · · = an = 1 Moreover, for n = 81, equality occurs when a1 = a2= · · · = a80=q3 3

4 and a81=√36

Remark 1 The inequality is not valid for n > 81 To prove this is is enough to let

a1= a2 = · · · = an−1= 1 and an= 2 in the homogeneous inequality

n a51+ a52+ · · · + a5n
2 ≥ a61+ a62+ · · · + a6n
a21+ a22+ · · · + a2n
2

We would get that (n − 1)(81 − n) ≥ 0, clearly false for n > 81

Remark 2 We can also prove the inequality in Proposition 1 using the Equal Variable Theorem ([2], [5]) According to the EV theorem, the following statement holds:

If 0 ≤ x1 ≤ x2 ≤ ≤ x81 such that x1+ x2+ · · · + x81= 81 and x

1 3

1 + x

1 3

2 + · · · + x

1 3

81

is constant, then the sum x

5 6

1 + x

5 6

2 + · · · + x

5 6

81 is minimal whenever x1 = x2= · · · =

x80≤ x81

Proposition 2 Let a1, a2, , an be positive real numbers such that

a1a2· · · an= 1

If m ≥ n − 1, then

am1 + am2 + · · · + amn + n(2m − n) ≥ (2m − n + 1) 1

a1 +

1

a2 + · · · +

1

an



Proof Since the case n = 2 and m = 1 is trivial, we may assume that m > 1 Let

ai= 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= 0 This inequality is equivalent to

f (x1) + f (x2) + · · · + f (xn) ≤ n · f x1+ x2+ · · · + xn

n

 ,

where f (u) = emu+ 2m − n − (2m − n + 1)e−u, u ∈ R.

We will prove that f (u) is right convex for u ≥ s, where s = x1 +x 2 +···+x n

f00(u) ≥ 0 for u ≥ 0 Taking the second derivative of f (u), we get

f00(u) = e−u[m2e(m+1)u− 2m + n − 1] > 0, because

m2e(m+1)u− 2m + n − 1 ≥ m2− 2m + n − 1 = (m − 1)2+ n − 2 > 0 According to the RCF theorem, it suffices to prove the inequality above for x2 =

x3 = · · · = xn≥ 0 or, equivalently, the original inequality for a2 = · · · = an≥ 1 :

xm+ (n − 1)ym+ n(2m − n) ≥ (2m − n + 1) 1

x +

n − 1 y



for m ≥ n − 1, 0 < x ≤ 1 ≤ y and xyn−1 = 1 Let us rewrite the inequality as

f (y) ≥ 0, where

ym(n−1) + (n − 1)ym+ n(2m − n) − (2m − n + 1)



yn−1+ n − 1

y



We have f0(y) = (n−1)g(y)ymn−m+1, g(y) = m(ymn−1) − (2m − n + 1)ymn−m−1(yn−1), and

g0(y) = ymn−m−2h(y), where

h(y) = m2nym+1− (2m − n + 1)[(m + 1)(n − 1)yn− mn + m + 1]

h0(y) = (m + 1)nyn−1[m2ym−n+1− (2m − n + 1)(n − 1)]

If m = n − 1 and n ≥ 3, then h(y) = n(n − 1)(n − 2) > 0 Otherwise, if m > n − 1 and n ≥ 2, then

m2ym−n+1− (2m − n + 1)(n − 1) ≥ m2− (2m − n + 1)(n − 1) = (m − n + 1)2 > 0, and hence h0(y) > 0 for y ≥ 1 Therefore, h(y) is strictly increasing on [1, ∞), and

h(y) ≥ h(1) = n[(m − 1)2+ n − 2] > 0 for y ≥ 1 Since h(y) > 0 implies g0y > 0, it follows that g(y) is strictly increasing

on [1, ∞) Then g(y) ≥ g(1) for y ≥ 1, and from ymn−m+1f0(y) = (n − 1)g(y) ≥ 0 it follows that f (y) is strictly increasing on [1, ∞)

Consequently, f (y) ≥ f (1) = 0 for y ≥ 1 For n = 2 and m = 1, the original inequality becomes an equality Otherwise, equlity occurs if and only if a1 = a2 =

· · · = an= 1

Remark 3 For m = n − 1, the following statement is true:

If a1, a2, , an are positive real numbers such that a1a2· · · an= 1, then

an−11 + an−12 + · · · + an−1n + n(n − 2) ≥ (n − 1) 1

a1

a2

+ · · · + 1

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 a1+ · · · + an

n



≥ (n − 1) (f (b1) + · · · + f (bn)) , where bj = n−11 Pn

j6=iaj, for all i

Consider the convex function f (x) = ex, replace a1, a2, , anwith (n −1) ln a1, (n − 1) ln a2, , (n − 1) ln an, and you get the desired result (see [4])

Remark 4 Replacing a1, a2, , anby x1

1,x1

2, ,x1

n the inequality in Proposition

2 becomes

1

xm1 +

1

xm2 + · · · +

1

xm n

+ (2m − n)n ≥ (2m − n + 1)(x1+ x + 2 + · · · + xn), where x1x2· · · xn = 1 We can also prove the inequality by the Equal Variable theorem:

If 0 < x1≤ x2 ≤ · · · ≤ xn such that

x1+ x2+ · · · + xn= constant and x1x2· · · xn= 1, then the sum x1m

1 +x1m

2 + · · · +x1m

n is minimal when 0 < x1= x2 = · · · = xn−1≤ xn

References

[1] V Cirtoaje, A generalization of Jensens Inequality, Gazeta Matematica Seria

A, 2 (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 Re-flections, 1 (2007).[http://reflections.awesomemath.org/2007 1.html]

[4] V Cirtoaje, Two Generalizations of Popovicius Inequality, Crux Math., 5 (2005), 313-318

[5] V Cirtoaje, The Equal Variable Method, Journal of Inequalities in Pure and Ap-plied Mathematics, 8 (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