Let $f\in\mathbb{R}[X_1,\ldots,X_n]$ be symmetric (or more generally invariant by a compact group). Then, there exists $n$ symmetric polynomials (more generally, $m$ $G$-invariant polynomial) - for example, the elementary symmetric polynomials $\sigma_1,\ldots,\sigma_n$ as mentioned by Bazin above, such that every symmetric polynomial can be (uniquely in the case of $G=S_n$) written as $f=g(\sigma_1,\ldots,\sigma_n)$. Now, the fact that $f$ is non-negative amounts to the same as saying that $g$ is non-negative on the image of the map $x\mapsto (\sigma_1,\ldots, \sigma_n)$. Procesi and Schwarz 1 -or in fact already earlier Procesi 2 in the case of symmetric polynomials, make the link to Hilbert's 17th problem: The image can be described as a basic semi-algebraic set, given by explicit inequalities, say $h_1\geq 0,\ldots, h_k\geq 0$. So one gets directly (see for example Theorem 2.1 in Procesi's paper) that $g=\sum_{1\leq i_1<i_2<\ldots<i_t\leq k} s_{i_1\cdots i_t} h_{i_1}h_{i_2}\cdots h_{i_t}$, where each $s_{i_1\cdots i_t}$ is a sum of squares of elementary symmetric functions. This is maybe the most general statement one can hope for, in the case of symmetric polynomials - and generally. Let me also add, that one can use in general representation theory of linear groups to get a good handle over the cones of symmetric ($G$-invariant) sums of squares of a given degree - in particular in the case of the symmetric group and more generally finite reflection groups, where one has a good control, since the representation in terms of the generators is unique. I Let me shamelessly also advertise my twothe papers on this topic here [3,4]invariant sums of squares decompositions - in the case of the symmetric group 3 and happily answer any questionin the (slightly) more general case 4.