Given a natural numbers $n$. Prove that $\left(\frac{\sum_{k=1}^{n}x_{k}}{n}\right)^{\sum\limits_{k=1}^{n}x_{k}}\leqq\prod_{k=1}^{n}x_{k}^{x_{k}}$ .

Question

Given a natural numbers $n$. Prove that $$\left ( \frac{\sum_{k= 1}^{n}x_{k}}{n} \right )^{\sum\limits_{k= 1}^{n}x_{k}}\leqq \prod\limits_{k= 1}^{n}x_{k}^{x_{k}}$$

I think that the "midway" is not solved easily. My only hint for my problem is the following inequality: $$\left ( \frac{n+ 1}{2} \right )^{\frac{n(n+ 1)}{2}}\leqq \prod\limits_{k= 1}^{n}k^{k}$$

Answer 1

For $f(x)=x\ln x$, (it is convex when $x\ge 0$) by Jensen Inequality,

$$nf(\dfrac{x_1+\dots +x_n}{n})\le f(x_1)+\dots +f(x_n) $$ Which is,

$$(x_1+\dots +x_n)\ln(\dfrac{x_1+\dots +x_n}{n})\le x_1\ln(x_1)+ \dots + x_n\ln(x_n) $$

Then taking exponential gives us,

$$\left(\frac{\sum_{k=1}^{n}x_{k}}{n}\right)^{\sum\limits_{k=1}^{n}x_{k}}\le\prod_{k=1}^{n}x_{k}^{x_{k}}$$