# An Inequality for Mixed Means

### Statement

Professor Dorin Marghidanu has posted the following problem and the solution at the CutTheKnotMath facebook page:

If $a_j\gt 0,$ $j=1,\ldots,m$ and $x_i\ge 0$, $i=1,\ldots,n.\;$ Then prove that:

$\displaystyle\begin{align} \sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}&\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}\\ &\ge\prod_{i=1}^{n}\prod_{j=1}^{m}\sqrt[mn]{a_j+x_i}. \end{align}$

### Solution

Define function $f:\;[0,\infty)$ as $f(x)=\sqrt[m]{(a_1+x)(a_2+x)\ldots (a_m+x)}.$ We have successively,

$\displaystyle f'(x)=\frac{1}{m}f(x)\left(\frac{1}{a_1+x}+\frac{1}{a_2+x}+\ldots+\frac{1}{a_m+x}\right)\gt 0,$

and

$\displaystyle\begin{align} f''(x)&=\frac{1}{m^2}f(x)\left(\frac{1}{a_1+x}+\frac{1}{a_2+x}+\ldots+\frac{1}{a_m+x}\right)^2\\ &\;\;-\frac{1}{m}f(x)\left(\frac{1}{(a_1+x)^2}+\ldots+\frac{1}{(a_m+x)^2}\right)\le 0, \end{align}$

by the Cauchy-Schwarz (often Cauchy-Bunyakovsky-Schwarz) inequality.

Hence function $f$ is *concave* and by Jensen's inequality for concave functions,

$\displaystyle f\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)\ge\frac{1}{n}\sum_{i=1}^{n}f(x_i).$

Specifically,

$\displaystyle \sqrt[m]{\left(a_1+\frac{1}{n}\sum_{i=1}^{n}x_i\right)\cdot\ldots\cdot\left(a_m+\frac{1}{n}\sum_{i=1}^{n}x_i\right)}\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{(a_1+x_i)\cdot\ldots\cdot (a_m+x_i)},$

or, in shorthand,

$\displaystyle \sqrt[m]{\prod_{j=1}^{m}\left(a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right)}\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}$

which proves the left inequality in the problem. The right inequality is the direct consequence of the AM-GM inequality:

$\displaystyle\begin{align} \frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}&\ge\sqrt[n]{\prod_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}}\\ &=\prod_{i=1}^{n}\sqrt[n]{\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}}\\ &=\prod_{i=1}^{n}\prod_{j=1}^{m}\sqrt[mn]{a_j+x_i}. \end{align}$

### References and a Remark

$\displaystyle G_m[a+A_n[x]]\ge A_n[G_m[a+x]]\ge G_n[G_m[a+x]]=G_m[G_n[a+x]].$

The inequality appeared in

- Dorin Marghidanu,
__Asupra unei inegalitati pentru medii compozite__,*Revista de matematica si informatica*, pp 3-5, Anul XIII, n. 2, April 2013.

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