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

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.

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