Prove that $\prod\limits_{i=1}^nx_i^{x_i}\ge\prod\limits_{i=1}^n{x_i^a}$

Question

Prove that $$\prod_{i=1}^nx_i^{x_i}\ge\prod_{i=1}^n{x_i^a}$$ for all $x_1,\dots,x_n\in\mathbb{R}^+$ where $a=\frac1n(x_1+\cdots+x_n)$.

Firstly, I've transformed the above inequality into the following form: $$x_1\ln x_1+\cdots+x_n\ln x_n\ge\frac1n(x_1+\cdots+x_n)(\ln x_1+\cdots+\ln x_n)$$ Now it is a bit hard to proceed. I've tried to apply mean inequlities $$Q_n\ge\cdots\ge Q_1\ge A\ge G\ge H$$ but none of them gave me useful result. After that I've tried Cauchy-Schwartz inequality, but the problem is because in that inequality square of product of sums is greater than square of sum of inner products, so I didn't find a way to apply it here. Any ideas?

Answer 1

Let $f(x)=x\ln{x}$.

Hence, $f$ is a convex function.

Thus, by Jensen and AM-GM $\sum\limits_{i=1}^nx_i\ln{x_i}\geq\sum\limits_{i=1}^nx_i\ln\frac{\sum\limits_{i=1}^nx_i}{n}\geq\sum\limits_{i=1}^nx_i\ln{\left(\prod\limits_{i=1}^nx_i\right)^{\frac{1}{n}}}=a\ln{\prod\limits_{i=1}^nx_i}$

and we are done!

Answer 2

This proof is incomplete one, so there is need of complete one. (I'm sorry). What is my idea is to approach it in integrals form. So what you have to prove is like this - $$\int_{0}^{T}x\ln x dx \geq \frac{1}{T}\int_{0}^{T}xdx \int_{0}^{T}\ln xdx$$

If $x_{1} < x_{2} < ... < x_{n}$, this would be a complete proof, and would be enough.

The inequilty in the above form will become the following if we do the integration. $$\frac{T^{2}\ln T}{2}-\frac{T^{2}}{4} \geq \frac{1}{T}\frac{T^{2}}{2}(T \ln T - T)$$ And doing some simplifications we will come to $$2\ln T -1 \geq 2 \ln T -2$$ Which is obvious.

Again. I'm sorry.