Minimizers of polynomial functions? Let $p: \mathbb{R}^2 \to \mathbb{R} $ be a polynomial function of two real variables. Suppose that
(P) $\; \; \; \; \; \; p(x,y) \ge 0, \; \; \forall \; (x,y) \in \mathbb{R}^2 $
1: Prove that if $p$ is of degree two, it has a minimizer over $\mathbb{R}$.
2: Prove that if a polynomial function $p$ verifies property (P) its degree is necessarily even.
3: Suppose that degree $p=2n$ with $n \gt 1  $. Does property (P) imply the existence of a minimizer for $p$ over $\mathbb{R}$?
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My ideas:
Since $f(x,y)$ is a degree two polynomial, it'd be of the form $$ f(x,y) = ax^2 +by^2 + cx + dy + kxy + l  $$
Taking all the second derivatives to compute the Hessian gives:
$$\left\lbrack
\matrix{2a & k\cr k & 2b}
\right\rbrack$$
My hunch is that since the function has to be equal or greater than zero, $a, b$ and $k$ have to be non-negative, which would mean the Hessian is always non-negative, which I think would mean that there has to be a minimizer.
I'm not really sure about the other parts, but I think that the answer to part "3" is YES. At least a local minimizer I think
 A: (1) and (2) are true, but (3) is false, and no Hessians are needed.
(1) is true : specializing (P) with $x=0$, we see that 
$by^2+dy+l\geq 0$ for any $y$. So either $b>0,d^2-4bl \leq 0$ or $b=d=l=0$.
In the latter case, we have $f(x,y)=ax^2+cx+kxy=x(ax+c+ky)$, so $f(1,y)=a+c+ky$
and (P) forces $k=c=0,a\geq 0$ ; then $0$ is a minimizer. 
So we can assume without loss that $b>0,d^2-4bl \leq 0$.
Similarly, interchanging $x$ and $y$, we can further assume $a>0,c^2-4al\geq 0$.
Next, we use the classic technique of "completing the square" ; i.e.
we write
$$
f(x,y)=ax^2+cx+b\bigg(\big(y+\frac{d+kx}{b}\big)^2-\big(\frac{d+kx}{b}\big)^2\bigg)+l=
g(x,Y)
$$
where $Y=y+\frac{d+kx}{b}$ and
$$
g(x,Y)=ax^2+cx-b\big(\frac{d+kx}{b}\big)^2+bY^2+l
$$
We see that $g$ satisfies $(P)$ just like $f$. So replacing $f$ with
$g$, we may assume  without loss that $d=k=0$. Similarly completing the square
in $x$, we can further assume that $c=0$. Then $f(x,y)=ax^2+by^2+l$ with $a,b$
nonnegative, and $l$ is a minimizer.
(2) is clear by specializing $(P)$ to $f(t,t)$, which is a univariate polynomial with a lower bound, so it must be of even degree.
(3) Counter-example : $f(x,y)=x^2+(xy-1)^2$. Note that 
$f(\frac{1}{n},n)=\frac{1}{n^2}$, so that $0$ is a lower bound, but it is never
attained.
