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 $$\displaystyle \prod^n_{i=1} (1+a_i^2) \le \cfrac {2^n}{n^{2n-2}}\left (\sum^n_{i=1} a_i\right)^{2n-2}$$


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 $(\clubsuit)$?
