does a convex polynomial always reach its minimum value? Consider a convex polynomial $p$ such that $p(x_1,~x_2,\dots x_n)\geq 0~\forall x_1,~x_2,\dots x_n\in \mathbb{R}^n$. Does the polynomial reach its minimum value?
This is not true for non-convex polynomials like $(1-x_1x_2)^2+x_1^2$, see the response of J.P. McCarthy on a similar question on general polynomials. This is not true for general functions either. Function
$e^x$ is infinitely differentiable, convex and bounded from below, but there is no $x$ that does reach the minimum value 0 (thanks to zhw for this simpler example).
 A: The restriction of $p$ to any straight line is a polynomial in one variable that is bounded below, therefore is either constant or goes to $\infty$ in both directions.  If there is a line $L$ on which $p$ is constant, then using convexity it is easy to see that $p$ must be constant on all lines parallel to $L$, and by taking a cross-section we reduce the dimension by $1$.  So we can assume wlog there is no line on which $p$ is constant. 
Now consider $A = \{x \in \mathbb R^n: p(x) < C\}$ where $C > p(0)$.
This is a convex set.  The restriction of $p$ to any ray through $0$ is a nonconstant polynomial in one variable and bounded below, therefore goes to $+\infty$ in both directions.  Thus $A$ contains no ray through $0$.  For each $s$ in the unit sphere $\mathbb S^{n-1}$, there is some $t > 0$ such that 
$p(ts) > C$  and by continuity this holds (with the same $t$) in some neighbourhood of $s$.  Note that by convexity, $p(t' s) > C$ for all $t' > t$.  Using compactness, we conclude that $A$ is bounded.  And then the infimum of $p$ is the infimum of $p$ on the compact set $\overline{A}$, which is attained.
EDIT: As requested, I'll expand on "using compactness".  For each $s \in \mathbb S^{n-1}$, there is $t > 0$ such that $p(ts) > C$.  Thus the open sets $\{s \in \mathbb S^{n-1}: p(t s) > C\}$ for $t > 0$ form an open covering of $\mathbb S^{n-1}$.  Because $\mathbb S^{n-1}$ is compact, this has a finite subcovering, i.e. $t_1, \ldots, t_k$ such that for every $s \in \mathbb S^{n-1}$, some $p(t_j s) > C$.   But that says 
$p(x) > C$ for all $x$ with $\|x\| \ge \max(t_1, \ldots, t_k)$, i.e. $\|x\| < \max(t_1, \ldots, t_k)$ for all $x \in A$.
