Problem. Let $n \in \mathbb{N}_{\ge 3}$. Let $a_1, a_2, \cdots, a_n > 0$ such that $\prod_{i=1}^n a_i = 1$. Prove that$$\prod^n_{i=1} (1+a_i^2) \le \frac {2^n}{n^{2n-2}}\left (\sum^n_{i=1} a_i\right)^{2n-2}.$$
I saw this problem recently which was originally asked about 12 years ago-- a Problem Statement Question (PSQ)which will be closed. There were six answers.Some users tackled the problem by the method of Lagrange Multipliers.I am afraid once it was closed, some users can not see the Q&A there.
Hope to see nice proofs.
As an alternative way, my approach is given below.
Let$$f(a_1, a_2, \cdots, a_n) := \frac {2^n}{n^{2n-2}}\left (\sum^n_{i=1} a_i\right)^{2n-2} - \prod^n_{i=1} (1+a_i^2).$$Consider the minimum of $f$ subject to$a_i > 0, \forall i$and $\prod_{i=1}^n a_i = 1$.It is easy to prove that the minimum exists (finite).
We claim that at minimum,among $a_1, a_2, \cdots, a_n$, there are at most two distinct values.Assume, for the sake of contraction, thatat minimum, there exists $a_i > a_j > a_k$.We have the following identity$$(1 + a_i^2)(1 + a_j^2)(1 + a_k^2)$$$$= (a_i + a_j + a_k - a_i a_j a_k)^2+ (a_i a_j + a_j a_k + a_k a_i - 1)^2.\tag{1}$$To proceed, we need the following Fact 1.
Fact 1. Let $p, r > 0; p^3 > 27r$ be given. Let $a, b, c > 0$ such that $a+b+c = p$ and $abc = r$. Then$ab + bc + ca$ is maximized only if $(a-b)(b-c)(c-a) = 0$,and $ab + bc + ca$ is minimized only if $(a-b)(b-c)(c-a) = 0$.
(The proof is easy. Actually, it is a result in the uvw method.)
By Fact 1 and (1), there exist $x, y > 0$ such that$a_i + a_j + a_k = 2x + y$, and $a_i a_j a_k = x^2y$,and $(1 + a_i^2)(1 + a_j^2)(1 + a_k^2)< (1 + x^2)(1 + x^2)(1 + y^2)$.Thus, if we replace $a_i, a_j, a_k$ with $x, x, y$,we have a strictly smaller objective value.This contradicts the optimality of $a_1, a_2, \cdots, a_n$.The claim is proved.
Thus, we can assume that $a_1 = a_2 = \cdots = a_k$,and $a_{k+1} = a_{k+2} = \cdots = a_n$ for some $k$. The problem is reduced to the following.
Problem 1. Let $x, y > 0$ and $1 \le k \le n$ and $x^k y^{n-k} = 1$. Prove that$$(1 + x^2)^k(1 + y^2)^{n-k}\le \frac{2^n}{n^{2n-2}}(kx + (n-k)y)^{2n-2}.$$