Dorin Marghidanu's Inequality in Many Variables

Solution 1

$\displaystyle \prod_{k=1}^{n}a_k = \prod_{k=1}^{n}\sqrt[n-1]{\prod_{i=1,i\ne k}^{n}a_i}\le \prod_{k=1}^{n}\frac{\displaystyle \sum_{i=1,i\ne k}^{n}a_i}{n-1}=\frac{\displaystyle\prod_{k=1}^{n}\sum_{i=1,i\ne k}^{n}a_i}{(n-1)^n}.$

Hence,

$\displaystyle \frac{\displaystyle\prod_{k=1}^{n}\sum_{i=1,i\ne k}^{n}a_i}{\displaystyle\prod_{k=1}^{n}a_k}\ge (n-1)^n$

which is equivalent to the required inequality. Equality holds when all $n$ variables are equal.

Solution 2

\displaystyle \begin{align} \prod_{k=1}^n\left[\frac{\displaystyle \sum_{i=1,i\neq k}^n a_i}{a_k}\right]^{\frac{1}{n}}&=\left[\prod_{k=1}^n\frac{\displaystyle \sum_{i=1,i\neq k}^n a_i}{a_k}\right]^{\frac{1}{n}} \\ &\geq\left[\prod_{k=1}^n\frac{\displaystyle (n-1)\left(\prod_{i=1,i\neq k}^n a_i\right)^{\frac{\displaystyle 1}{n-1}}}{a_k}\right]^{\frac{1}{n}}~\text{(AM-GM)}\\ &=\left[\frac{\displaystyle (n-1)^n\left(\prod_{i=1}^n a_i\right)^{\frac{n-1}{n-1}}}{\displaystyle \prod_{k=1}^n a_k}\right]^{\frac{1}{n}} \\ &=(n-1). \end{align}

Solution 3

Tinkering before exact proof. Math is fun!

$\displaystyle \frac{\displaystyle \sum_{i=1}^na_i-a_k}{n-1}\ge\frac{\displaystyle\left(\prod_{i=1}^na_i\right)^{\frac{1}{n-1}}}{a_k^{\frac{1}{n-1}}}.$

From here,

\displaystyle \begin{align} \prod_{k=1}^{n}\left(\frac{\displaystyle \sum_{i=1}^na_i-a_k}{a_k}\right)^{\frac{1}{n}}&\ge\prod_{k=1}^n\left( \frac{\displaystyle (n-1)\left(\prod_{i=1}^na_i\right)^{\frac{1}{n-1}}}{a_k^{\frac{1}{n-1}+1}}\right)^{\frac{1}{n}}\\ &=\prod_{k=1}^n\left((n-1)a_k^{-\frac{1}{n-1}-1}\left(\prod_{i=1}^na_i\right)^{\frac{1}{n-1}}\right)^{\frac{1}{n}}\\ &=(n-1)\prod_{k=1}^na_k^{\frac{-\frac{1}{n-1}-1}{n}}\left(\prod_{i=1}^na_i\right)^{\frac{1}{(n-1)n}}=n-1. \end{align}

Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu with a solution of his (Solution 1). Solution 2 is by Amit Itagi; Solution 3 is by N. N. Taleb.