I want to show that for all $x_i > 0$:
$$\sum_{i=1}^{n}\dfrac{x_i}{x_1 + \ldots+x_n}\;\log(x_i) \geq \log\left(\dfrac{x_1 + \ldots + x_n}{n}\right)$$
I thought of Jensen's inequality but since $\log$ is a concave function, this seems to give me the opposite as for $x_i > 0, a_i \geq 0$, it holds: $a_1\log(x_1) + \ldots + a_n\log(x_n) \leq \log(a_1x_1 + \ldots + a_nx_n)$.
Or is there some algebraic manipulation we can do on the original inequality such that we can use Jensen (or any other inequality for that matter)?
Best Answer
The inequality can be rewritten to $$ \frac{x_1 + \ldots + x_n}{n} \cdot \log\left(\frac{x_1 + \ldots + x_n}{n}\right) \le \frac 1n \sum_{i=1}^n x_i \log(x_i) \, , $$ and that is true because the function $f(x) = x \log(x)$ is convex.