Prove that $ (\frac{\sum_{i=1}^n x_i}{n})^{\sum_{i=1}^n x_i} \le \prod_{i=1}^n {x_i}^{x_i}$ $, \forall x_i>0, n\ge1 $

Question

Prove that $ (\frac{\sum_{i=1}^n x_i}{n})^{\sum_{i=1}^n x_i} \le \prod_{i=1}^n {x_i}^{x_i}$ $, \forall x_i>0, n\ge1 $
(The second sum in the left-hand side of the inequality is an exponent)

I've been trying to solve this for a day now, mostly trying to use Jensen's, but I can't seem to figure out how to intertwine the $n$ in the left-hand side with the other variables.

Answer 1

The hint.

Use Jensen for the convex function $f(x)=x\ln{x}.$

Answer 2

Oop, found an answer... I guess writing it down in an adequate manner (like the MathJax display) helped me see the "link" between the left- and right-hand side

The inequality is equivalent to $$ (\frac{\sum_{i=1}^n x_i}{n})^{(\frac{\sum_{i=1}^n x_i}{n}) * n } \le \prod_{i=1}^n {x_i}^{x_i} $$ which is then equivalent to, logarithmizing, $$ n* f(\frac{\sum_{i=1}^n x_i}{n}) \le \sum_{i=1}^n f(x_i) $$ where $ f(x)=x*ln(x) $, and since $f:(0, +\infty) \to \mathbb R$ is convex on its domain, Q.E.D

(Thanks @Michael Rozenberg, too! )