About the inequality $\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}$ Hi It's a generalization found on Aops starting from this question Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$ (see comments) :
Let $x_i>0$, $1\leq i\leq n$,$n\geq 3$ such that $x_1=x_{n+1}$ then we have :
$$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}\tag{I}$$
At first glance it seems that Am-Gm is too weak so as in my answer I use Radon's inequality (see vivid edit) we need to show :
$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_{i+1}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq n\sqrt{2}$$
Then I used Minkowski's inequality with $p=\frac{2}{3}$ we need to show :
$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}+\left(\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq n\sqrt{2}$$
Edit :
We can apply Jensen's inequality and we need to show  :
$$\frac{2^{-\frac{1}{2}}\left(\sum_{i=1}^{n}\left(\frac{1}{x_{i}}\right)^{\frac{2}{3}}+\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_{i}}+x_{i}\right)}}-n\sqrt{2}\geq 0$$
Question :
How to achieve my work (if true) or show $(I)$?
 A: Lemma 1: For $a,b>0$,
$$2\sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
Proof:
$$\left(\frac{1}{\sqrt{a}}-\sqrt{b}\right)^2 \geqslant 0$$
$$\frac{1}a + b  \geqslant  2\frac{\sqrt{b}}{\sqrt{a}}$$
$$\left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)^2= \frac{1}a + b + 2\frac{\sqrt{b}}{\sqrt{a}} \leqslant 2\left(\frac{1}a + b\right)  $$
and the result follows.

Lemma 2: For $a>0$,
$$\sqrt{\frac{1}{a}+a} \leqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{a}-1\right)$$
Proof:
$$(\sqrt{a}-1)^4\geqslant 0$$
$$a^2-4\sqrt{a}a+6a-4\sqrt{a}+1\geqslant 0$$
$$2(a^2-2\sqrt{a}a+3a-2\sqrt{a}+1)\geqslant a^2+1$$
$$2(a-\sqrt{a}+1)^2\geqslant a^2+1$$
$$2\left(\sqrt{a}-1+\frac{1}{\sqrt{a}}\right)^2\geqslant a+\frac{1}{a}$$
and the result follows.

Lemma 3:
For $a,b>0$,
$$\frac{{\frac{1}{a}+b}}{\sqrt{\frac{1}{a}+a}} \geqslant \sqrt2 (\sqrt{b}-\sqrt{a}+1)$$
Proof:
By AM-GM and then Lemma 1:
$$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\sqrt{\frac{1}{a}+a} \geqslant 2 \sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
Combining with Lemma 2:
$$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+ \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{a}-1\right)  \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
and the result follows.

Theorem: For $x_i>0$,
$$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}$$
Proof:
Let $a=x_i$ and $b=x_{i+1}$ in Lemma 3 and sum.
