# 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.

## 2 Answers

The hint.

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

• Thank you very much, just figured out the solution! – Parallelism Alert Feb 9 at 14:53
• @Parallelism Alert You are welcome! – Michael Rozenberg Feb 9 at 14:54

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! )