Let $x_1, x_2, \dots, x_n > 1$ and $x_1 x_2 \dots x_n=n+1$. Prove that$$\left(\frac{1}{1^2(x_1-1)}+x_1\right)\left(\frac{1}{2^2(x_2-1)}+x_2\right)\cdots\left(\frac{1}{n^2(x_n-1)}+x_n\right)\geq \frac{(n+1)(n+2)}{2}$$Euality holds when $x_1=2,x_2=\frac{3}{2},x_3=\frac{4}{3} ,\cdots,x_n=\frac{n+1}{n}.$

Stuttgarden wrote: Let $n$ be a positive integer. Let $x_1, x_2, \dots, x_n > 1$ be real numbers whose product is $n+1$. Prove that \[\left(\frac{1}{1^2(x_1-1)}+1\right)\left(\frac{1}{2^2(x_2-1)}+1\right)\cdots\left(\frac{1}{n^2(x_n-1)}+1\right)\geq n+1\]and find for which values equality holds. Euality holds when $x_1=2,x_2=\frac{3}{2},x_3=\frac{4}{3} ,\cdots,x_n=\frac{n+1}{n}.$

Since $x_k>1$, then \[1+\frac1{k^2(x_k-1)}=\frac{(k+1)^2}{k^2x_k}+\frac{(x_kk-k-1)^2}{k^2x_k(x_k-1)}\ge\frac{(k+1)^2}{k^2x_k},\]So \[\prod_{k=1}^n\left[1+\frac1{k^2(x_k-1)}\right]\ge\prod_{k=1}^n\frac{(k+1)^2}{k^2x_k}=\frac{(n+1)^2}{\prod\limits_{k=1}^nx_k}=n+1.\]

By CS we have: $$(\frac{1}{i^2 (x_i - 1)} + 1)( x_i - 1 + 1) \geq (\frac{1}{i \sqrt{x_i - 1}} \cdot \sqrt{x_i - 1} + 1)^2 = (\frac{i+1}{i})^2$$Multiplying them all we have that: $$(n+1) \cdot \prod (\frac{1}{i^2 (x_i - 1)} + 1) = \prod (\frac{1}{i^2 (x_i - 1)} + 1) \prod x_i \geq \prod (\frac{i+1}{i})^2 = (n+1)^2$$So we are done.

By AM-GM we have \[ \frac{1+1+\dots +1+k(x_k-1)}{k+1}\ge \sqrt[k+1]{k(x_k-1)} \Leftrightarrow \frac{k}{k+1}x_k\ge \sqrt[k+1]{k(x_k-1)} \]Multiplying over $k$ we get \[ 1 = \frac{\prod x_k}{n+1}\ge \prod \sqrt[k+1]{k(x_k-1)}\qquad (*) \]Using again AM-GM \[ \frac{1}{k}+\frac{1}{k}+\dots +\frac{1}{k} + \frac{1}{k^2(x_k-1)}\ge (k+1)\sqrt[k+1]{\frac{1}{k^{k+2}}\frac{1}{x_k-1}} = \frac{k+1}{k}\frac{1}{\sqrt[k+1]{k(x_k-1)}} \]Multiplying over $k$ we get \[ \left(\frac{1}{1^2(x_1-1)}+1\right)\left(\frac{1}{2^2(x_2-1)}+1\right)\cdots\left(\frac{1}{n^2(x_n-1)}+1\right)\ge \frac{n+1}{\prod \sqrt[k+1]{k(x_k-1)}} \]Using $(*)$ we finally get \[ \left(\frac{1}{1^2(x_1-1)}+1\right)\left(\frac{1}{2^2(x_2-1)}+1\right)\cdots\left(\frac{1}{n^2(x_n-1)}+1\right)\ge n+1 \]It is easy to deduce that equality only holds when $x_1=2,x_2=\frac{3}{2},x_3=\frac{4}{3} ,\cdots,x_n=\frac{n+1}{n}.$

Given, $\prod_{i=1}^n x_i = n+1$, we have $$\prod_{i=1}^n\left(\frac{1}{i^2(x_i-1)}+1\right)\stackrel{\text{T2}}{\geq}\prod_{i=1}^n\frac{(i+1)^2}{i^2x_i}=\frac{(n+1)^2}{\prod_{i=1}^n x_i}=n+1.$$Equality is achieved when $i^2(x_i-1)=k_i$ and $i^2=k_ii$, or in other words, when $x_i=\frac{i+1}{i}$, which can easily be verified.

By T2's lemma we have $$\frac{1}{i^2(x_i-1)}+1=\frac{1}{i^2(x_i-1)}+\frac{i^2}{i^2}\geq \frac{(1+i)^2}{i^2(x_i-1)+i^2}=\frac{(i+1)^2}{i^2x_i}$$Equality holds when $1/i^2(x_i-1)=i/i^2\iff x_i=1+1/i$. Multiplying cyclically yields the desired inequality.