$(\frac{s-a_1}{n-1})^{a_1} \cdot (\frac{s-a_2}{n-1})^{a_2}\cdots (\frac{s-a_n}{n-1})^{a_n} \le (\frac{s}{n})^s$

Question

If $a_1,a_2,\ldots,a_n$ be $n$ positive rational numbers and $s=a_1+a_2+\cdots+a_n$, prove that $$\left(\frac{s-a_1}{n-1}\right)^{a_1} \cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n} \le \left(\frac{s}{n}\right)^s$$

My try:

Consider $\left(\frac{s-a_1}{a_1}\right),\left(\frac{s-a_2}{a_2}\right),\ldots,\left(\frac{s-a_n}{a_n}\right)$ be $n$ numbers with associated weights $a_1,a_2,\ldots,a_n$ .

Then by Weighted AM-GM inequality

$$\frac{a_1\cdot\left(\frac{s-a_1}{a_1}\right)+a_2\cdot\left(\frac{s-a_2}{a_2}\right)+\cdots+a_n\cdot\left(\frac{s-a_n}{a_n}\right)}{a_1+a_2+\cdots+a_n}\ge \left[\left(\frac{s-a_1}{a_1}\right)^{a_1}\cdot \left(\frac{s-a_2}{a_2}\right)^{a_2}\cdots \left(\frac{s-a_n}{a_n}\right)^{a_n}\right]^{\frac{1}{a_1+a_2+\cdots+a_n}}$$

Therefore

$$\left(n-1\right)^{a_1+a_2+\cdots+a_n}\ge \left(\frac{s-a_1}{a_1}\right)^{a_1}\cdot \left(\frac{s-a_2}{a_2}\right)^{a_2}\cdots \left(\frac{s-a_n}{a_n}\right)^{a_n}$$

$$\left(\frac{s-a_1}{n-1}\right)^{a_1}\cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n}\le a_1^{a_1}\cdot a_2^{a_2}\cdots a_n^{a_n}$$

I am stuck. please give me some hint.Thanks

Answer 1

let $f(x)=x\ln\frac{a-x}{n-1},$ where $0<x<a$.

Thus, $$f''(x)=\frac{x-2a}{(x-a)^2}<0,$$ which says that your inequality it's just Jensen.

Indeed, since $f$ is a concave function, we obtain: $$\sum_{i=1}^na_i\ln\frac{s-a_i}{n-1}\leq n\cdot\frac{\sum\limits_{i=1}^na_i}{n}\ln\frac{s-\frac{\sum\limits_{i=1}^na_i}{n}}{n-1}=s\ln\frac{s}{n}.$$

Answer 2

AM-GM gives: $\frac{\left(\frac{s-a_1}{n-1}\right)a_1 + \left(\frac{s-a_2}{n-1}\right)a_2+ \cdots +\left(\frac{s- a_n}{n-1}\right)a_n}{a_1 + a_2 +\cdots+a_n} \ge \left(\left(\frac{s-a_1}{n-1}\right)^{a_1} \cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n}\right)^{\frac{1}{s}}$

LHS $= \frac {2\sum\limits_{1\le i\lt j\le n} a_ia_j}{(n-1)s}$

We can show AM of the numbers $\frac{s}{n} \ge \frac {2\sum\limits_{1\le i\lt j\le n} a_ia_j}{(n-1)s}$

$\implies (n-1)s^2 \ge 2n\sum\limits_{1\le i\lt j\le n} a_ia_j$

$\implies (n-1)\sum\limits_{i=1}^n a_i^2 \ge 2\sum\limits_{1\le i\lt j\le n} a_ia_j$

which is true since we can write $\binom{n}{2}$ equations of the form:

$ a_i^2 + a_j^2 \ge 2 a_i a_j$ and add them up.