Let $x_1, x_2, \dots, x_n$ be positive numbers such that $$\frac{1}{x_1^2}+\frac{1}{x_2^2}+\cdots+\frac{1}{x_n ^2}=\frac{1}{n}$$
Denote $x_{n+1}=x_1$, show that `$$\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}+x_1x_2+x_2x_3+\dots+x_{n}x_{n+1}\geq n^3+1$$
I want to apply Cauchy-Schwarz's inequality to show this, I also try to show the case for $n=3$, but I still have no clue.
Here is my work:
It follow from A.M.-G.M inequality that
$$
\sum_{i=1}^n x_ix_{i+1}
=\sum_{i=1}^n
\frac{1}{\frac{1}{x_{i}x_{i+1}}}
\geq\sum_{i=1}^n\frac{2}
{\left(\frac{1}{x_i}\right)^2
+\left(\frac{1}{x_{i+1}}\right)^2}$$
By the given condition and Cauchy-Schwarz's inequality gives the following.
$$
\begin{align}
&\frac{1}{n}\sum_{i=1}^n x_ix_{i+1}\\
&=\sum_{i=1}^n \frac{1}{x_i^2}\sum_{i=1}^n x_ix_{i+1}\\
&\geq\sum_{i=1}^n
\left(\frac{1}{x_i^2}+\frac{1}{x_{i+1}^2}\right)
\sum_{i=1}^n
\frac{1}{\left(\frac{1}{x_i}\right)^2+\left(\frac{1}{x_{i+1}}\right)^2}\\
&\geq n^2
\end{align}
$$
That is, $\sum_{i=1}^n x_ix_{i+1}\geq n^3$.
But I find that the remaining part of L.H.S. has maximum $1$ by Cauchy-Schwarz's inequality again, so this cannot give the desired result.
Best Answer
let $\sqrt[n]{\prod\limits_{cyc}x_1}=x$.
Thus, by AM-GM from the condition we obtain: $$\frac{1}{n}=\sum_{cyc}\frac{1}{x_1^2}\geq\frac{n}{x^2},$$ which gives $x\geq n$.
Now, by AM-GM again $$\sum_{cyc}\frac{1}{x_1}+\sum_{cyc}x_1x_2=\sum_{cyc}\frac{1}{x_1}+n^3\cdot\frac{\sum\limits_{cyc}x_1x_2}{n^3}\geq(n^3+1)\sqrt[n^3+1]{\sum_{cyc}\frac{1}{x_1}\left(\frac{\sum\limits_{cyc}x_1x_2}{n^3}\right)^3}\geq$$ $$\geq(n^3+1)\sqrt[n^3+1]{\frac{n}{x}\left(\frac{x^2}{n^2}\right)^3}=(n^3+1)\sqrt[n^3+1]{\frac{x^5}{n^5}}\geq n^3+1.$$