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?


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!

  • $\begingroup$ On which numbers did you apply Jensen? $\endgroup$ – ultramath Dec 14 '16 at 23:05
  • $\begingroup$ @ultramath $x_1$, $x_2$,..., $x_n$. $\endgroup$ – Michael Rozenberg Dec 14 '16 at 23:12
  • $\begingroup$ As I know, Jensen's inequality is $\lambda_1f(x_1)+\cdots+\lambda_nf(x_n)\ge f(\lambda_1x_1+\cdots+\lambda_nx_n)$ where lambdas are nonnegative numbers which sum is $1$ and $f$ is convex function. So, 1) What is your lambdas here? 2) What is your $f$ here? 3) Did you prove that sum of lambdas is $1$? 3) How did you get from step 1 to step 2 (why did you assume that $\ln x_i\ge \ln a$)? $\endgroup$ – ultramath Dec 15 '16 at 0:07
  • $\begingroup$ $f(x)=x\ln{x}$, $\lambda_1=\lambda_1=...=\frac{1}{n}$. Id est, $\frac{x_1\ln{x_1}+...+x_n\ln{x_n}}{n}\geq\frac{x_1+...+x_n}{n}\ln\frac{x_1+...+x_n}{n}$. $\endgroup$ – Michael Rozenberg Dec 15 '16 at 0:32
  • $\begingroup$ Ok, I got it. The only thing which had confused me is because you hadn't write $\frac1n$ on both sides. Now it is clear. $\endgroup$ – ultramath Dec 15 '16 at 14:11

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.

  • $\begingroup$ This has absolutelly nothing to do with my question. Numbers $x_1,\dots,x_n$ are arbitrary positive real numbers, but you integrated it like they are consecutive. It is just a coincidence that your integral works, but actually it cannot help us at all. $\endgroup$ – ultramath Dec 14 '16 at 19:17
  • $\begingroup$ No you are wrong. Only if there are repetition among $x_{i}$-s only than the proof is incomplete. If you want to do something with it, you can even use Dirac's function to come to your case. $\endgroup$ – kolobokish Dec 14 '16 at 19:21
  • $\begingroup$ No, I'm not wrong. This is exactly what I've already said. Your "proof" fails because $x_i$ may be distanct from $x_j$ for some $i,j\in\{1,\dots,n\}$. $\endgroup$ – ultramath Dec 14 '16 at 19:23

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.