Let $p(x)$ be a real polynomial that is bounded below. Prove that there is a real number $x_0$ such that $p(x) ≥ p(x_0)$ for all $x$.

Question

Let $p(x)$ be a real polynomial that is bounded below. Prove that there is a real number $x_0$ such that $p(x) ≥ p(x_0)$ for all $x$.

This is listed on some practice problems for a contest math group I'm in. The question seems really easy but I'm not sure how to solve it.

One thought I had would be to decompose $\Bbb R$ into a countably infinite union of closed intervals, and then apply the extreme value theorem to each of those, and finally to order the infimum using the axiom of choice (inside the set because it is closed) and choose the smallest one (which may be repeated). But I don't know if this is correct or elegant.

For the countable union of closed intervals, I could take the set {... ,[0,1] [1,2], [2,3], [3,4], ...} which would cover $\Bbb R$, and I believe is countable. Hints appreciated.

It's also worth mentioning that with calculus you could describe all the types of intervals that are possible in a polynomial of degree $n$ and proceed from there, but I haven't fleshed that out either.

Answer 1

Hint: $p$ has finitely many critical points. Consider a compact interval containing them.

Answer 2

Let $b$ be the glb for for $f(\mathbb R)$. Since $f$ is a polynomial, it follows easily from the hypotheses, that $\lim_{x\to \pm \infty}=\infty$, so there is an interval $[-n,n]$ such that $f(x)>b$ whenever $x\in [-n,n]^c.$ Since $[-n,n]$ is compact, $f$ attains a minimum on $[-n,n]$, which in fact, is easily seen to be $b$