show this inequality $\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\frac{1}{x_{i}}+n\right)^n$

We need to prove that $$\frac{\sum\limits_{k=1}^nx_k+n}{\sum\limits_{k=1}^n\frac{1}{x_k}+n}\geq\sqrt[n]{\prod\limits_{k=1}^nx_k}$$ and by the Vasc's EV Method (see here:

http://emis.ams.org/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf ,

Corollary 1.7(a))

it's enough to prove our inequality for $$x_1=b^n\leq x_2=...=x_n=a^n$$

where $a\geq1$ and $b\geq1$ or $$\frac{(n-1)a^n+b^n+n}{\frac{n-1}{a^n}+\frac{1}{b^n}+n}\geq a^{n-1}b$$ or $$ab^{n-1}\left((n-1)a^n-na^{n-1}b+b^n\right)\geq a^n-nab^{n-1}+(n-1)b^n.$$ Let $a=xb$.

Hence, $x\geq1$ and we need to prove that $$b^nx\left((n-1)x^n-nx^{n-1}+1\right)\geq x^n-nx+n-1$$ and since $b\geq1$ and by AM-GM $$(n-1)x^n-nx^{n-1}+1\geq0,$$ it's enough to prove that $$(n-1)x^{n+1}-nx^n+x\geq x^n-nx+n-1$$ or $f(x)\geq0$, where $$f(x)=(n-1)x^{n+1}-(n+1)x^n+(n+1)x-(n-1).$$ Now, $$f'(x)=(n+1)(n-1)x^n-n(n+1)x^{n-1}+n+1$$ and $$f''(x)=n(n+1)(n-1)x^{n-1}-(n-1)n(n+1)x^{n-2}\geq0.$$ Thus, $$f'(x)\geq f'(1)=0$$ and from here $$f(x)\geq f(1)=0$$ and we are done!