This looks like a stupid question, but the obvious answer (if there is one) eludes me …
Let $f(x,y)$ be a symmetric polynomial in $x$ and $y$. Then $f$ attains a minimum $m$ on the compact set $K=[0,1]^2$. Is is true that there is always at least one point $M(x,y) \in K$ on the diagonal (i.e. with $x=y$) such that $f(M)=m$ ?
Update 20:00 To make the question slightly less stupid, let us ask if there is always one point on the diagonal or the boundary of $K$ where the minimum is attained ?
The symmetric polynomial function
$$f(x,y)=\left[\left(x-\frac{1}{3}\right)^2+\left(y-\frac{2}{3}\right)^2\right]\cdot\left[\left(x-\frac{2}{3}\right)^2+\left(y-\frac{1}{3}\right)^2\right]$$
attains its minima precisely when it is zero, which occurs precisely when either $(x,y)=(1/3,2/3)$ or $(2/3,1/3)$, both of which occur inside $K=[0,1]^2$ but not on the diagonal or boundary.