<s>[INST] Prove that $\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge (n^2-1)^n$ Let $n$ be a positive integer and $x_1,...,x_n$ positive reals such that $x_1+...+x_n=1$. 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.
 [/INST] Lemma:$$\left(\dfrac{1}{x}-x\right)\left(\dfrac{1}{y}-y\right)\ge 4^3\left(\dfrac{z^3}{3^9\cdot x^2y^2}\right)^{\frac{1}{4}}~~~~~(n=3)$$
$$
LHS=\dfrac{1}{xy}\left(\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}+xy\right)\left(x+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)\left(y+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)\ge RHS$$
then
$$\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{y^2}-1\right)\ge 4^3\left(\dfrac{z^3}{3^9\cdot x^2y^2}\right)^{\frac{1}{4}}\cdot\dfrac{1}{xy}$$
for any postive $n$ we have  use AM-GM same methods
$$\left(\dfrac{1}{x^2_{1}}-1\right)\left(\dfrac{1}{x^2_{2}}-1\right)
\ge \left(n-\dfrac{1}{n}\right)^2\left(\dfrac{1}{n}\right)^{\dfrac{n^3-3n^2+2n-2}{(n-1)^2(n+1)}}\cdot(x_{1}x_{2}\cdots x_{n})^{\frac{n}{(n-1)^2}}\left(\dfrac{1}{x_{1}x_{2}}\right)^{\frac{n^2+1}{n^2-1}}\cdot\dfrac{1}{x_{1}x_{2}}$$
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