# An Inequality with Constraint V

### Statement

Leonard Giugiuc, Dan Sitaru
18 December, 2015

Let $0\lt x_k,$ $k=1,\dots,n$ such that $\displaystyle\sum_{k=1}^{n}x_k=1,$ $n\ge 1.$ Then

$\displaystyle\prod_{k=1}^{n}x_k^{1/x_k}\le \frac{1}{n^{n^2}}.$

### Solution

Without loss of generality, assume $x_1\le x_2\le\ldots\le x_n$ so that $\displaystyle\frac{1}{x_1}\ge \frac{1}{x_2}\ge\ldots\ge\frac{1}{x_n}.$ By the AM-GM inequality, $1=\sum_{k=1}^{n}x_k\ge n\sqrt[n]\prod_{k=1}^{n}x_k,$ implying $\displaystyle\prod_{k=1}^{n}x_k\le\frac{1}{n^n}\lt 1.$

Since function $f(x)\log_{a}x,\;$ with $0\lt a\lt 1,\;$ is monotone decreasing,

$\log_{x_1x_2\ldots x_n}x_1\ge\log_{x_1x_2\ldots x_n}x_2\ge\ldots...\ge\log_{x_1x_2\ldots x_n}x_n.$

Therefore, by the Chebyshev and Jensen inequalities,

\begin{align} \sum_{k=1}^{n}\frac{1}{x_k}\log_{x_1x_2\ldots x_n}x_k&\ge\frac{1}{n}\left(\sum_{k=1}^{n}\log_{x_1x_2\ldots x_n}x_k\right)\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\\ &\ge\frac{1}{n}\log_{x_1x_2\ldots x_n}x_1x_2\ldots x_n\cdot\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\\ &\ge\frac{1}{n}\cdot 1\cdot\frac{n^2}{x_1+x_x+\ldots+x_n}\\ &\ge n. \end{align}

It follows that

$\displaystyle\log_{x_1x_2\ldots x_n}\left(x_1^{\displaystyle\frac{1}{x_1}}\cdot\ldots\cdot x_n^{\displaystyle\frac{1}{x_n}}\right)\ge n=\log_{x_1x_2\ldots x_n}(x_1x_2\ldots x_n)^n,$

which is equivalent to

$\displaystyle x_1^{\displaystyle\frac{1}{x_1}}\cdot\ldots\cdot x_n^{\displaystyle\frac{1}{x_n}}\le (x_1x_2\ldots x_n)^n\le\left(\frac{x_1+\ldots+x_n}{n}\right)^{n^2}=\frac{1}{n^{n^2}}.$

This is the required inequality.