# Optimize a symmetric polynomial on a compact set

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 ?

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$f=1-(x-y)^2$ gets smaller the farther away from the diagonal you are, so it's a counterexample. – Lopsy Dec 18 '12 at 18:40
Trivial exemple. Let's a polinom $p(x,y)$ convex such that $p(0,0)=0$. Then your answer is yes. – MathOverview Dec 18 '12 at 18:40
And $-x(1-x)y(1-y)(x-y)^2$ has no minimum on the diagonal or on the boundary. – WimC Dec 18 '12 at 18:51
It seems unlikely - you should be able to start with an asymmetric polynomial $g$ that has a minimum at, say, $(\frac{1}{4}, \frac{3}{4})$ and symmetrize by setting $f(x,y) = g(x,y)+g(y,x)$. – Steven Stadnicki Dec 18 '12 at 18:52

$$f(x,y)=\left[\left(x-\frac{1}{3}\right)^2+\left(y-\frac{2}{3}\right)^2\right]\cdot\left[\left(x-\frac{2}{3}\right)^2+\left(y-\frac{1}{3}\right)^2\right]$$
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.