An Inequality with Exponents from a Calculus Lemma II

Lemma

For $u\gt 0,\;$ $v\in (0,1),\;$ the following inequality holds:

$\displaystyle u^v\gt\frac{u}{u+v}.$

Proofs of the lemma and a list of additional applications have been placed on a separate page.

Proof 1

Under the conditions of the statement, $\displaystyle\frac{a+b}{2},\frac{b+c}{2},\frac{c+a}{2}\in (0,1),$ making the application of Lemma possible:

$\displaystyle \left(\frac{a+b}{2}\right)^c\gt\frac{\displaystyle\frac{a+b}{2}}{\displaystyle\frac{a+b}{2}+c}.$

Hence, $a+b+2c\gt 2^c(a+b)^{1-c}.\;$ Similarly,

$a+2b+c\gt 2^b(c+a)^{1-b}\;$ and
$2a+b+c\gt 2^a(b+c)^{1-a}.$

By adding the three we obtain

$4(a+b+c)\gt 2^a(b+c)^{1-c}+2^b(c+a)^{1-b}+2^c(a+b)^{1-c},$

the required inequality.

Generalization

Assume $a_i\in (0,1),\;$ $i=0,1,\ldots,n-1.$ Then

$\displaystyle\sum_{cycl}2^{a_i}(a_{i+1}+a_{i+2})^{1-a_i}\lt 4\sum_{cycl}a_i,$

where indices are taken modulo $n.$

Proof 2

First note that

$\displaystyle 2^a(b+c)^{1-a}=2\left(\frac{b+c}{2}\right)^{1-a}=\left(1-\frac{2-b-c}{2}\right)^{1-a},$

with $\displaystyle 0\lt\frac{2-b-c}{2}\lt 1,\;$ which allows to use Bernoulli's reverse inequality

$(1-x)^y\le 1-xy,\;$ for $x,y\in (0,1).$

Thus we successively obtain

\displaystyle\begin{align} \left(1-\frac{2-b-c}{2}\right)^{1-a} &\le 1-\frac{2-b-c}{2}(1-a)\\ &=1-\left(1-\frac{b+c}{2}\right)(1-a)\\ &=1-1+\frac{b+c}{2}-\frac{b+c}{2}\cdot a\\ &=\frac{b+c}{2}+a-\left(\frac{b+c}{2}\right)a. \end{align}

So we have

$\displaystyle \left(\frac{b+c}{2}\right)^{1-a} \le \frac{b+c}{2}+a-\frac{b+c}{2} a.$

Similarly,

\displaystyle\begin{align} \left(\frac{c+a}{2}\right)^{1-b} \le \frac{c+a}{2}+b-\frac{c+a}{2}\cdot b\\ \left(\frac{a+b}{2}\right)^{1-c} \le \frac{a+b}{2}+c-\frac{a+b}{2}\cdot c\\ \end{align}

\displaystyle\begin{align} \sum_{cycl}2^a(b+c)^{1-a} &= 2\sum_{cycl}\left(\frac{b+c}{2}\right)^{1-a}\\ &\le 2\sum_{cycl}\left[\frac{b+c}{2}+a-\left(\frac{b+c}{2}\right)a\right]\\ &=2\sum_{cycl}\frac{b+c}{2}+2\sum_{cycl}a-2\sum_{cycl}\frac{b+c}{2}a\\ &=4\sum_{cycl}a-\sum_{cycl}a(b+c)\\ &\lt 4\sum_{cycl}a. \end{align}

Acknowledgment

The problem, with the above solution, has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. It was previously published as Problem 169 in the Missouri Journal of Mathematical Sciences, v 22, n 1.

Proof 2 is by Marian Dinca.