show this inequality $\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\frac{1}{x_{i}}+n\right)^n$ Let $x_{i}\ge 1$,show that
$$\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n\right)^n$$
or 
$$\left(\dfrac{\sum_{i=1}^{n}x_{i}+n}{\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n}\right)^n\ge \prod_{i=1}^{n}x_{i}$$
and it seem use AM-GM inequality?
$$\sum_{i=1}^{n}x_{i}\ge n\sqrt[n]{x_{1}x_{2}\cdots x_{n}}$$
$$\sum_{i=1}^{n}\dfrac{1}{x_{i}}\ge \dfrac{n}{\sqrt[n]{x_{1}x_{2}\cdots x_{n}}}$$ let $\sqrt[n]{x_{1}x_{2}\cdots x_{n}}=t$,since 
$$\Longleftrightarrow \left(\dfrac{t+1}{\frac{1}{t}+1}\right)^n\ge t^n$$But I can't it 

we have to prove  :
$$\left(\dfrac{\sum_{i=1}^{n}x_{i}+n}{\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n}\right)^n\ge \prod_{i=1}^{n}x_{i}$$ 
Let $x_i\geq 1$ be real numbers so we have :
$$\frac{1}{\sum_{i=1}^{n}S_i}\left(\sum_{i=1}^{n}S_i[(-x_i+\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n})(n)x_i^{n-1}+x_i^n]\right)\ge \prod_{i=1}^{n}x_{i}$$ 
Where :
$$S_i=\frac{\prod_{i=1}^{n}x_{i}}{|((-x_i+\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n})(n)x_i^{n-1}+x_i^n)-1|}$$
And : 
$$S_{min}=\frac{n\prod_{i=1}^{n}x_{i}}{|((-x_{min}+\frac{(\sum_{i=1}^{n}x_{i})+n}{(\sum_{i=1}^{n}\frac{1}{x_{i}})+n})(n)x_{min}^{n-1}+x_{min}^n)-1|}$$
After that we use the theorem 5 of this link :
You just have to replace the function by $\phi(x)=x^n$ with domain $[x_{min};x_{max}]$ and $d=\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n}$ and remark that we have :
$x_{min}\leq d \leq x_{max}$ and put $S_i=W_i$ to get the LHS.
