Symmetric rational function is a rational function in the elementary symmetric polynomials.

Let $f\in \mathbf{Q}(T_1,\ldots,T_n)$ be a symmetric rational function, then we can write $f=\frac{g}{h}$ with $g,h\in \mathbf{Z}[T_1,\ldots,T_n]$.

If $h$ is a symmetric polynomial, then $g=hf$ is symmetric too. By the fundamental theorem of symmetric polynomials, $f$ is a rational function in the elementary symmetric polynomials.

If $h$ is not a symmetric polynomial, consider $h'=\prod_{\sigma\in S_n\setminus \{e\}} \sigma h$. Then $hh'$ is a symmetric polynomial by construction. By writing $f=\frac{g}{h}=\frac{gh'}{hh'}$, we reduce ourselves to the previous case.

For $\sigma \in S_n$, let $\lambda_\sigma \in K^\times$ be such that $\sigma(g) = \lambda_\sigma g$ and $\sigma(h) = \lambda_\sigma h$. Then $\sigma \mapsto \lambda_\sigma$ is a group homomorphism $\lambda : S_n \to K^\times$. (When $n \geq 6$, there are at most two such homomorphisms.)

We want to show that $\lambda$ is trivial. Suppose not, and let $(i, j)$ be a transposition that does not lie in the kernel of $\lambda$. We will arrive at a contradiction with the fact that $g$ and $h$ are coprime. Because $\lambda_{(i, j)} \neq 1$, we obtain that $g(T_1, \ldots, \widehat T_j, T_i, \ldots, T_n) = 0$. (We substituted $T_i$ for $T_j$.) When $g$ is considered as a polynomial over $K((T_k)_{k \neq j})$, this means that $T_i$ is a root of $g$. Thus $g$ is divisible by $(T_i - T_j)$. Similarly, $h$ is divisible by $(T_i - T_j)$. But by Gauss's lemma, $g$ and $h$ are coprime in $K((T_k)_{k \neq j})[T_j]$, a contradiction.