Inequality $(1+x_1)(1+x_2)\ldots(1+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}\right)\geq 2n^2.$ Let $n\geq 2$, and $x_1,x_2,\ldots,x_n>0$. Show that $$(1+x_1)(1+x_2)\ldots(1+x_n)\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots+\dfrac{1}{x_n}\right)\geq 2n^2.$$
For $n=2$, this reduces to $(1+x_1)(1+x_2)(x_1+x_2)\geq 8x_1x_2$. We may apply the Arithmetic-Geometric mean inequality on each of the term on the left to get the result.
However, when $n\geq 3$, this doesn't work anymore.
[Source: Ukrainian competition problem]
 A: Just another way:
$$\begin{align}
\left(\prod_{k=1}^n(1+x_k) \right) \left(\sum_{k=1}^n\frac1{x_k}\right) &= \left(\prod_{k=1}^n \left(\underbrace{\frac1{n-1}+\frac1{n-1}+\cdots+\frac1{n-1}}_{n-1 \text{ times}}+x_k \right) \right) \left(\sum_{k=1}^n\frac1{x_k}\right) \\
&\ge \left(\frac1{\sqrt[n+1]{(n-1)^{n-1}}}+\frac1{\sqrt[n+1]{(n-1)^{n-1}}}+\cdots+\frac1{\sqrt[n+1]{(n-1)^{n-1}}} \right)^{n+1} \\
&= \left(\frac{n}{\sqrt[n+1]{(n-1)^{n-1}}} \right)^{n+1} = \frac{n^{n+1}}{(n-1)^{n-1}} 
\end{align}$$
where the inequality used is Holder's.  So it is enough to show that 
$$n^{n-1}\ge 2(n-1)^{n-1} \iff \left(1+\frac1{n-1}\right)^{n-1} \ge 2$$
which follows from Bernoulli's inequality for $n\ge 2$.
A: Applying AM-GM to $\frac{1}{x_1}+\cdots+\frac{1}{x_n}$ yields
$$
\frac{1}{x_1}+\cdots+\frac{1}{x_n}\geq n\frac{1}{\sqrt[n]{x_1\cdots x_n}}.
$$
Next, apply AM-GM again to each $(1+x_i)$ term:
$$
1+x_i=\frac{1}{n-1}+\cdots+\frac{1}{n-1}+x_i\geq n\sqrt[n]{\frac{x_i}{(n-1)^{n-1}}}.
$$
Putting the above results together, we note that it suffices to prove the following inequality
$$
\frac{n^{n+1}}{(n-1)^{n-1}}\geq 2n^2\iff\left(\frac{n}{n-1}\right)^{n-1}\geq 2\iff\left(1+\frac{1}{n-1}\right)^{n-1}\geq 2.
$$
But the last inequality above is just the Bernoulli's inequality. The claim follows.
