Denote the power-sum (symmetric) polynomials by $p_k(a;n)=\sum_{i=1}^na_i^k$; where $a=(a_1,\dots, a_n)$ and also we need the Hadamard product $ab=(a_1b_1,\dots,a_nb_n)$. Write $p_k(a)$ for $p_k(a;n)$ when there is no confusion.
Remark. Observe that $p_2(a)p_2(b)-p_2(ab)=\sum_{i\neq j}a_i^2b_j^2\geq0$. We prove $$[p_2(a)p_2(b)+p_1(ab)^2-p_2(ab)]^2-p_2(a^2)p_2(b^2)\geq0.\tag1$$ Induct on $n$. The case $n=2$ is noted as trivial. Assume true for $n$, we show for $n+1$; i.e. $$[(x^2+p_2(a))(y^2+p_2(b))+(xy+p_1(ab))^2-x^2y^2-p_2(ab)]^2-(x^4+p_2(a^2))(y^4+p_2(b^2))\geq0.\tag2$$ After expansion, the LHS of (2) is a polynomial in even powers of $x$ and $y$. It suffices to study the following coefficients (the others follow by symmetry):
$[x^0y^0]$: the is exactly the induction assumption, hence positive.
$[x^2y^0]$: $2p_2(b)[p_1(ab)^2+p_2(a)p_2(b)-p_2(ab)]\geq0$, by remark from above.
$x^4y^0]$: $p_2(b)^2-p_2(b^2)\geq0$, by remark.
$[x^2y^2]$: $6p_1(ab)^2+4p_2(a)p_2(b)-2p_2(ab)\geq0$, by remark.
$[x^4y^2]$: $2p_2(b)\geq0$.
$x^0y^4]$$[x^0y^4]$: p_2(a)^2-p_2(a^2)\geq0$$p_2(a)^2-p_2(a^2)\geq0$, by remark.
Therefore, the claim (1) is valid for all $n$.
The intent of this method is to reveal that the inequality (1) is not as sharp as we may wish.