# Dan Sitaru's Cyclic Inequality In Many Variables

### Solution

By the AM-GM inequality:

\displaystyle\begin{align} &\frac{a^5+(a^5+b^5+c^5+d^5)}{5}\ge\sqrt[5]{a^5\cdot a^5b^5c^5d^5}=a\cdot abcd,\\ &\frac{b^5+(a^5+b^5+c^5+d^5)}{5}\ge\sqrt[5]{b^5\cdot a^5b^5c^5d^5}=b\cdot abcd,\\ &\frac{c^5+(a^5+b^5+c^5+d^5)}{5}\ge\sqrt[5]{c^5\cdot a^5b^5c^5d^5}=c\cdot abcd,\\ &\frac{d^5+(a^5+b^5+c^5+d^5)}{5}\ge\sqrt[5]{d^5\cdot a^5b^5c^5d^5}=d\cdot abcd. \end{align}

$\displaystyle\frac{5(a^5+b^5+c^5+d^5)}{5}\ge (a+b+c+d)abcd,$

which is exactly the required inequality.

### Generalization

Prove that, for integer $n\gt 0\,$ and $a_k,\,k\in\overline{1,n},$

$\displaystyle \left(\sum_{k=1}^na_k\right)\left(\prod_{k=1}^na_k\right)\le \sum_{k=1}^na_k^{n+1}.$

Written this way, the problem begs for further generalization:

Prove that, for integer $n,t\gt 0\,$ and $a_k,\,k\in\overline{1,n},$

$\displaystyle \left(\sum_{k=1}^na_k^t\right)\left(\prod_{k=1}^na_k\right)\le \sum_{k=1}^na_k^{n+t}.$

Proof

By the AM-GM inequality,

\displaystyle\begin{align} \sum_{j=1}^n\left(\frac{\displaystyle ta_{j}^{n+t}+\sum_{k=1}^na_k^{n+t}}{n+t}\right)&\ge\sum_{j=1}^n\sqrt[n+t]{a_j^{t(n+t)}\cdot\prod_{k=1}^na_k^{n+t}}\\ &=\sum_{j=1}^na_j^t\cdot\prod_{k=1}^na_k\\ &=\left(\sum_{j=1}^na_j^t\right)\cdot\left(\prod_{k=1}^na_k\right). \end{align}

It remains only to note that

\displaystyle\begin{align} \sum_{j=1}^n\left(\frac{\displaystyle ta_{j}^{n+t}+\sum_{k=1}^na_k^{n+t}}{n+t}\right)&=\frac{\displaystyle t\sum_{j=1}^na_j^{n+t}+\sum_{j=1}^n\sum_{k=1}^na_k^{n+t}}{n+t}\\ &=\frac{\displaystyle t\sum_{j=1}^na_j^{n+t}+n\sum_{k=1}^na_k^{n+t}}{n+t}\\ &=\frac{\displaystyle (n+t)\sum_{k=1}^na_k^{n+t}}{n+t}\\ &=\sum_{kj=1}^na_k^{n+t}. \end{align}

### Acknowledgment

Dan Sitaru has shared a problem from the Romanian Mathematical Magazine, with a beautiful solution by Kunihiko Chikaya. Both the problem and the solution suggest a generalization.