2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

The hint.

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

$\endgroup$
  • $\begingroup$ Thank you very much, just figured out the solution! $\endgroup$ – Parallelism Alert Feb 9 at 14:53
  • $\begingroup$ @Parallelism Alert You are welcome! $\endgroup$ – Michael Rozenberg Feb 9 at 14:54
2
$\begingroup$

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! )

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.