TITLE: A more elementary proof that if $x_i>0$ for $1\leq i\leq n$, and $\sum x_i=1$, then $(x_1+\frac{1}{x_1})\cdots(x_n+\frac{1}{x_n})\geq(n+\frac1n)^n$
QUESTION [6 upvotes]: For $x_i>0$, $1\leq i\leq n$ and $\sum_i x_i=1$, show that
$$\left(x_1+\frac{1}{x_1}\right)\cdots \left(x_n+\frac{1}{x_n}\right)\geq \left(n+\frac{1}{n}\right)^n$$
I think this can be proved easily by Jensen's inequality. However, my child commented that they have not learned Jensen's inequality. I am wondering whether there is a more "elementary" proof using, for instance, only AM-GM or Cauchy, or maybe by induction(?). Many thanks.

REPLY [3 votes]: First let $a_i,b_i\geq 0$ and use AM-GM:
$$n = \sum_{i=1}^n\dfrac{a_i}{a_i+b_i}+\sum_{i=1}^n\dfrac{b_i}{a_i+b_i}\geq \dfrac{n(\sqrt[n]{a_1a_2\dots a_n}+\sqrt[n]{b_1b_2\dots b_n})}{\sqrt[n]{\prod_{i=1}^n(a_i+b_i)}}$$
or equivalently we obtain:
$$\prod_{i=1}^n(a_i+b_i)\geq (\sqrt[n]{a_1a_2\dots a_n}+\sqrt[n]{b_1b_2\dots b_n})^n\quad (1).$$
Take $a_i = x_i$ and $b_i = \frac{1}{x_i}$ and we have:
$$\prod_{i=1}^n\left(x_i+\frac{1}{x_i}\right)\geq\left(r+\frac 1r\right)^n\quad (2)$$
where $x_1x_2\dots x_n = r^n.$ Since $x_i$ sum up to $1$, AM-GM readily gives $r\leq \frac 1n$ which then yields:
$$(r-n)(rn-1)\geq 0\iff r+\frac 1r\geq n+\frac 1n\quad (3)$$
and combining $(2)$ and $(3)$ immediately concludes the problem in an elementary fashion.