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]

Solution 1:

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.

Solution 2:

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 $n\ge 2$.