An Inequality problem relating $\prod\limits^n(1+a_i^2)$ and $\sum\limits^n a_i$ 
Let $(a_1,\space a_2,\space \cdots, \space a_n) \in \mathbb R^n_+$ such that $\displaystyle \prod^n_{i=1 }a_i = 1$. Prove that $$\displaystyle \prod^n_{i=1} (1+a_i^2) \le \cfrac {2^n}{n^{2n-2}}\left (\sum^n_{i=1} a_i\right)^{2n-2}$$

 A: As a partial answer, consider the polynomial:
$$P(x)=\prod_{k=1}^{n}(x+a_k)=\sum_{j=0}^{n}x^{n-j} \binom{n}{j} S_j(a_1,\ldots,a_n),$$
where $\binom{n}{j} S_j(a_1,\ldots,a_n)$ is the $j$-th elementary symmetric polynomial in the variables $(a_1,\ldots,a_n)$. By hypothesis we have $S_n=1$. Moreover:
$$ (\clubsuit)\quad P(i)P(-i)=\prod_{k=1}^{n}(1+a_k^2)=\left(\sum_{j=0}^{\lfloor n/2 \rfloor}(-1)^j\binom{n}{2j}S_{2j}\right)^2+ \left(\sum_{j=0}^{\lfloor (n-1)/2 \rfloor}(-1)^j\binom{n}{2j+1}S_{2j+1}\right)^2,$$
and by Mac Laurin's inequality we have
$$ S_1\geq S_2^{1/2}\geq\ldots\geq S_n^{1/n}=1.$$
Are someone able to derive
$$ P(i)P(-i)\leq 2^n\,S_1^{2n-2} $$
from $(\clubsuit)$?
A: I have not solved the problem completely, but I have found how to reduce it to an easier-looking problem, in two variables instead of $n$. I use  the Lagrange multiplier method. Perhaps someone else can fill the blanks.
Let $\Omega=\lbrace (x_1,x_2, \ldots ,x_n) | x_i >0 \ (1 \leq i\leq n) \rbrace$, and let $M>0$. Consider the following optimization problem on $\Omega$ :
Maximize $f (x_1,x_2, \ldots ,x_n)$ subject to $g(x_1,x_2, \ldots ,x_n)=h(x_1,x_2, \ldots ,x_n)=0$, where
$$
\begin{array}{lcl}
f (x_1,x_2, \ldots ,x_n) &=& (1+x_1^2)(1+x_2^2) \ldots (1+x_n^2) \\
g(x_1,x_2, \ldots ,x_n) &=& x_1+x_2+ \ldots +x_n-M \\
h(x_1,x_2, \ldots ,x_n) &=& x_1x_2 \ldots x_n-1 \\
\end{array}
$$
The set $K_M=\lbrace (x_1,x_2, \ldots ,x_n) | g(x_1,x_2, \ldots ,x_n)=h(x_1,x_2, \ldots ,x_n)=0\rbrace$ is closed and bounded in the open set $\Omega$. So $f$ attains a minimum on $K_M$ at a point $p$, and we then know that Lagrange multipliers exist, i.e. there are two constants $\lambda$ and $\mu$ such that 
$$
\frac{\partial f}{\partial x_k}(p)=\lambda \frac{\partial g}{\partial x_k}(p)+\mu \frac{\partial h}{\partial x_k} \ (1 \leq k \leq n) 
$$
In other words,
$$
(1) \ 2x_k\prod_{j\neq k}(1+x_j^2)=\lambda+\mu(\prod_{j\neq k}x_j)  \ (1 \leq k \leq n) 
$$
Now (1) is a linear system of $n$ equations in two unknowns $\lambda$ and $\mu$. If $i$ and $j$ are indices such that $x_j \neq x_i$, using the $i$-th and the $j$-th equation we can solve for $\lambda$ and $\mu$ :
$$
(2) \  \lambda=\frac{2(x_i+x_j)}{(x_i^2+1)(x_j^2+1)}\frac{A}{B}, \ \  \ \mu=\frac{x_ix_j(x_ix_j-1)}{(x_i^2+1)(x_j^2+1)}\frac{A}{B} \ (\text{where} \ A=\prod_{j=1}^{n}(1+x_j^2), B= \prod_{j=1}^{n}x_j)
$$
Since (2) must hold for all pairs of indices $i,j$ with $x_i \neq x_j$, we deduce that either $\lbrace x_k \rbrace$ has at most two elements, or $n=3$ and $x_1x_2+x_1x_3+x_2x_3=1$.
Now this latter case is impossible since it would imply
$$
(x_1-x_2)^2+(x_1-x_3)^2+(x_2-x_3)^2=2((x_1+x_2+x_3)^2-(x_1x_2+x_1x_3+x_2x_3
))=0
$$
and hence all the $x_i$ would be equal to $1$ (as $x_1x_2x_3=1$),contradicting  $x_1x_2+x_1x_3+x_2x_3=1$.
So $\lbrace x_k \rbrace$ consists of at most two elements $a$ and $b$ (with possibly $a=b$). Denoting the number of occurrences by $n_a$ and $n_b$, we thus have $n_a+n_b=n$ and 
$$
\begin{array}{l}
f(x_1, \ldots ,x_n)=(1+a^2)^{n_a}(1+b^2)^{n_b}, \\
1=x_1x_2x_3 \ldots x_n=a^{n_a}b^{n_b}, \\
 x_1+x_2+ \ldots+x_n=n_aa+n_bb 
\end{array}
$$
It seems that new ideas are needed at this point to finish the solution.
A: Let $l(a_1,\dots,a_n) = \prod\limits^n_{i=1} (1+a_i^2)$
Let $r(a_1,\dots,a_n) = \cfrac {2^n}{n^{2n-2}}\left (\sum\limits^n_{i=1} a_i\right)^{2n-2}$
$\cfrac{d^nl}{da_1\dots da_n}( a_1,\dots,a_n ) = 2 ^ n\prod\limits^n_{i=1} a_i$
$\cfrac{d^nr}{da_1\dots da_n}( a_1,\dots,a_n ) = \cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!}(\sum\limits^n_{i=1} a_i)^{n-2}$
$(\sum\limits^n_{i=1} a_i)^{n} \ge n^n \prod\limits^n_{i=1} a_i$
$(\sum\limits^n_{i=1} a_i)^{n-2} \ge n^{n-2} (\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}}$
$\cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!}(\sum\limits^n_{i=1} a_i)^{n-2} \ge \cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!} n^{n-2} (\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}} = \cfrac {2^n}{n^{n}}\cfrac {(2n-2)!}{(n-2)!}(\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}}$
$1 \le \cfrac{1}{n^{n}}\cfrac {(2n-2)!}{(n-2)!}$
After this line, I assume $(\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}} = 1$ which I can't according to the comments.
$2^n \le \cfrac {2^n}{n^{n}}\cfrac {(2n-2)!}{(n-2)!} \le \cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!}(\sum\limits^n_{i=1} a_i)^{n-2}$
$\cfrac{d^nl}{da_1\dots da_n}( a_1,\dots,a_n ) \le \cfrac{d^nr}{da_1\dots da_n}( a_1,\dots,a_n )$
$l( a_1,\dots,a_n ) \le r( a_1,\dots,a_n )$
$\prod\limits^n_{i=1} (1+a_i^2) \le \cfrac {2^n}{n^{2n-2}}\left (\sum\limits^n_{i=1} a_i\right)^{2n-2}$
A: I think I have a serious way to solve this problem :
The idea is to use the function $arcth(x)$ and the relation $arcth(x)+arcth(y)=arcth\Big(\dfrac{x+y}{1+xy}\Big)$
So we put :
$a_i=arcth(b_i)$
We get :
$$\sum_{i=1}^{n}a_i=\sum_{i=1}^{n}arcth(b_i)=arcth(\frac{\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}S_{2k+1}(b_1,\cdots,b_n)}{\sum_{k=0}^{\lfloor n/2 \rfloor}S_{2k}(b_1,\cdots,b_n)})$$
Where $S_k$ are the $k^{th}$ symmetric polynomial.
Now the idea is to use the property of functions exponentially convex :

Exponentially convex function $\phi:(a,b)→(0,∞)$ is log-convex :
  $$\phi(\frac{x+y}{2})\leq \sqrt{\phi(x)\phi(y)} \quad\forall x,y \in(a,b)$$

Or equivalently we have after some substitution :
$$\phi(ln(\sqrt{uv}))\leq \sqrt{\phi(ln(u))\phi(ln(v))} $$
Or by extension  :
$$\phi(ln(\sqrt{\prod_{i=1}^{n}x_i}))\leq \sqrt{\prod_{i=1}^{n}\phi(ln(x_i))} $$
The function $\phi$ wich interest us is :
$\phi(x)=e^{\sqrt{e^{x}-1}}$ and of course the $x_i$ are :
$x_i=(a_i^2+1)=(arcth(b_i)^2+1)$
So we get for the RHS :
$$\sqrt{\prod_{i=1}^{n}\phi(ln(x_i))}=e^{\sum_{i=1}^{n}arcth(b_i)}=\prod_{i=1}^{n}\sqrt{\frac{x+1}{1-x}}=\sqrt{\frac{x+y}{x-y}} $$
If we put $y=\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}S_{2k+1}(b_1,\cdots,b_n)$ and $x=\sum_{k=0}^{\lfloor n/2 \rfloor}S_{2k}(b_1,\cdots,b_n)$
So it remains to prove :
$$\sqrt{\frac{A+1}{1-A}}\leq e^{\sqrt{\frac{2^n}{n^{2n-2}}arcth(A)^{2n-2}-1}}$$
With $A=\frac{y}{x}$ on the interval $[tanh(n);1[$ (wich correspond to the initial condition of the inequality and see the comment of Pedro Tamaroff if you don't see what I mean )
Wich is obvious . 
