Let $n$ be a positive integer and $x_1,…,x_n$ positive reals such that $x_1+\dots+x_n=1$. Prove that $$\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge (n^2-1)^n.$$
Now notice that if we apply Jensen's Inequality to $$f(x)=\ln\left(\frac{1}{x_i^2}-1\right)$$ We would get the exact inequality…except that $f$ is not convex on $(0,1)$. But I do know this,
Claim: If for all $k\le n$ we have $0< a_k\le 1/2$. Then, $$\prod_{i=1}^n\left(\frac{1}{a_i}-1\right)\ge \left(\frac{n}{a_1+…+a_n}-1\right)^n$$
The proof is not too hard. Just consider $$g(x)=\ln\left(\frac{1}{x}-1\right)$$
$g$ is convex on $(0,1/2)$ and you can finish with Jensen. Now what if we let $a_i=x_i^2$? Well, we would get $$\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge \left(\frac{n}{x_1^2+…+x_n^2}-1\right)^n\ge (n^2-1)^n$$
Because $x_1+…+x_n=1$ implies, $$\sum_{i=1}^n x_i^2\ge \frac{1}{n}\left(\sum_{i=1}^n x_i\right)^2=\frac{1}{n}$$
And we're done…Or are we? The argument is only valid if $x_i^2\le 1/2$ for all $i$. Well if there exists $x_i^2>1/2$ Can we still get the inequality?
By the way there is a way to solve the inequality without any use of Jensen. But I want to know if we can solve it like that.
Best Answer
WLOG, assume that $x_1 \le x_2 \le \cdots \le x_n$.
We split into two cases:
Case 1: $x_n < \frac{1}{\sqrt 3}$
Note that $x \mapsto \ln(\frac{1}{x^2} - 1)$ is convex on $(0, \frac{1}{\sqrt 3})$. We have $$\sum_{i=1}^n \ln\left(\frac{1}{x_i^2} - 1\right) \ge n\ln\left(\frac{1}{(\frac{x_1 + x_2 + \cdots + x_n}{n})^2} - 1\right) = n\ln(n^2 - 1).$$
Case 2: $x_n \ge \frac{1}{\sqrt 3}$
Let $y_1 = x_1, y_2 = x_2, \cdots, y_{n-2} = x_{n-2}$ and $y_{n-1} = y_n = \frac{x_{n-1} + x_n}{2}$. Then $0 < y_i < \frac{1}{2}$ for all $i$.
Note that $x \mapsto \ln(\frac{1}{x^2} - 1)$ is convex on $(0, \frac{1}{\sqrt 3})$. We have \begin{align*} \sum_{i=1}^n \ln\left(\frac{1}{y_i^2} - 1\right) \ge n\ln\left(\frac{1}{(\frac{y_1 + y_2 + \cdots + y_n}{n})^2} - 1\right) = n\ln(n^2 - 1) \end{align*} which results in $$\prod_{i=1}^{n-2}\left(\frac{1}{x_i^2}-1\right) \cdot \left(\frac{1}{(\frac{x_{n-1} + x_n}{2})^2} - 1\right)^2\ge (n^2-1)^n$$
It remains to prove that $$\left(\frac{1}{x_{n-1}^2}-1\right)\left(\frac{1}{x_n^2}-1\right) - \left(\frac{1}{(\frac{x_{n-1} + x_n}{2})^2} - 1\right)^2 \ge 0$$ which is true since \begin{align*} &\left(\frac{1}{a^2}-1\right)\left(\frac{1}{b^2}-1\right) - \left(\frac{1}{(\frac{a + b}{2})^2} - 1\right)^2 \\[6pt] ={}& \frac{a^2 + 6ab + b^2 - (a^2 + 4ab + b^2)(a + b)^2}{a^2b^2(a+b)^4}(a-b)^2\\ \ge{}& 0 \end{align*} where $a = x_{n-1}, b = x_n$ (using $a + b < 1$).
We are done.