algebraic geometry – Commutation of integral closure and group invariance

We work over $mathbb{C}$. Let $G$ be a reductive group acting on a normal affine variety $X$: we have an induced $G$-action on $k(X)$ given by
$$gcdot p(x)=p(g^{-1}cdot x).$$
Thanks to the GITwe know the good quotient $Y$ is affine and given by $Y=text{Spec} k(X)^G$.

I would like to prove that $Y$ is still normal, that is $k(X)^G=k(Y)=overline{k(Y)}=overline{k(X)^G}$ (where the closure is taken in the field $k(X)$, that is, the operation of integral closure and $G$-invariance is commutative

The inclusion $k(X)^Gsubset overline{k(X)^G}$ is obvious, so let us prove the reverse one. By the normality, it suffices to show that

$$overline{k(X)^G}subset overline{k(X)}^G,$$
I take $fin overline{k(X)^G}$, so that I can write

$$f^n(x)+a_{1}(x)f^{n-1}(x)+ldots+a_{n-1}(x)f(x)+a_n(x)=0$$
with $a_iin k(X)^G=overline{k(X)}^G$. The $G$-invariance allow me to see that the above formula works also
$$f^n(g^{-1}cdot x)+a_{1}(g^{-1}cdot x)f^{n-1}(g^{-1}cdot x)+ldots+a_{n-1}(g^{-1}cdot x)f(g^{-1}cdot x)+a_n(g^{-1}cdot x)=0$$
for any $gin G$, but now I’m a bit stuck as I don’t know how to continue (I can subsract them both but I don’t see what this imply).