A more elementary proof that if $x_i>0$ for $1\leq i\leq n$, and $\sum x_i=1$, then $(x_1+\frac{1}{x_1})\cdots(x_n+\frac{1}{x_n})\geq(n+\frac1n)^n$ 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.
 A: Yes, Jensen kills it.
There is also the following way.
Let $x_i=\frac{a_i}{n}$.
Thus, $$\sum_{i=1}^na_i=n$$ and we need to prove that:
$$\sum_{i=1}^n\left(\ln\left(\frac{a_i}{n}+\frac{n}{a_i}\right)-\ln\left(n+\frac{1}{n}\right)\right)\geq0$$ or
$$\sum_{i=1}^n\left(\ln\left(\frac{a_i}{n}+\frac{n}{a_i}\right)-\ln\left(n+\frac{1}{n}\right)+\frac{n^2-1}{n^2+1}(a_i-1)\right)\geq0,$$  which is true because for any $x>0$ we have $f(x)\geq0,$ where
$$f(x)=\ln\left(\frac{x}{n}+\frac{n}{x}\right)-\ln\left(n+\frac{1}{n}\right)+\frac{n^2-1}{n^2+1}(x-1).$$
Indeed, $$f'(x)=\frac{(x-1)((n^2-1)x^2+2n^2x+n^4+n^2)}{(n^2+1)(n^2+x^2)x},$$
which gives $x_{min}=1$ and we are done!
A: 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.
A: This proof is probably the most elementary one that I can come up with. (However, I have never been trained in olympiad-style inequalities, so it is quite likely that a much easier solution exists.)
We first prepare two lemmas:

Lemma 1. If $0 < x \leq y \leq 1$, then $x+\frac{1}{x} \geq y+\frac{1}{y}$.
Proof. This immediately follows from $ \bigl( x+\frac{1}{x} \bigr) - \bigl( y+\frac{1}{y} \bigr) = \frac{(y-x)(1-xy)}{xy} $.
Lemma 2. Let $a_1, a_2, b_1, b_2 \in (0, 1]$ satifsy $a_1 + a_2 = b_1 + b_2$ and $|b_1 - b_2| \leq |a_1 - a_2|$. Then
$$ \left(a_1 + \tfrac{1}{a_1}\right)\left(a_2 + \tfrac{1}{a_2}\right) \geq \left(b_1 + \tfrac{1}{b_1}\right)\left(b_2 + \tfrac{1}{b_2}\right).$$
Proof. By writing $ a_1 a_2
= \bigl(\frac{a_1+a_2}{2}\bigr)^2 - \bigl(\frac{a_2-a_1}{2}\bigr)^2 $ and $ b_1 b_2
= \bigl(\frac{b_1+b_2}{2}\bigr)^2 - \bigl(\frac{b_2-b_1}{2}\bigr)^2 $, we get $a_1 a_2 \leq b_1 b_2$. Then by Lemma 1,
\begin{align*}
\left(a_1 + \tfrac{1}{a_1}\right)\left(a_2 + \tfrac{1}{a_2}\right)
&= \left( a_1 a_2 + \tfrac{1}{a_1 a_2} \right) + \tfrac{(a_1 + a_2)^2}{a_1 a_2} - 2 \\
&\geq \left( b_1 b_2 + \tfrac{1}{b_1 b_2} \right) + \tfrac{(b_1 + b_2)^2}{b_1 b_2} - 2 \\
&= \left(b_1 + \tfrac{1}{b_1}\right)\left(b_2 + \tfrac{1}{b_2}\right).
\end{align*}

Now returning to the original problem, let $n\geq 2$ and $x_1, \dots, x_n > 0$ be such that $\sum_{k=1}^{n} x_k = 1$. This forces that $x_k \leq 1$ for all $k$.
Suppose that not all of $x_i$'s are equal to $\frac{1}{n}$. Then there must exist $i$ and $j$ such that $x_i < \frac{1}{n} < x_j$. Then we can find $x^*_i$ and $x^*_j$ such that
$$ x_i + x_j = x^*_i + x^*_j, \qquad |x^*_i - x^*_j| \leq |x_i - x_j|, \qquad \text{and}\qquad \tfrac{1}{n} \in \{ x^*_i, x^*_j \}. $$
So, replacing $x_i$ and $x_j$ by $x^*_i$ and $x^*_j$ respectively, we can lower both the number of terms different from $\frac{1}{n}$ and the value of the products $\prod_{k=1}^{n}(x_k + \frac{1}{x_k})$. Therefore, by repeating this procedure at most $n$ times, we obtain the desired lower bound.
