A simple proof that non negative polynomials have even degree I am looking for a simple proof that a non-negative polynomial in n variables has always even degree. I have proved it but using Artin's Theorem ( Every non-negative polynomial is the sum of squares of rational functions), but I think that there are simpler proofs.
I know that this question could be off-topic, but I couldn't get answers at MSE.
 A: This can be proved via an elementary argument.
Univariate case
Let's first look at the univariate case. Suppose $f(x) = x^{2r+1} + \sum_{0 \le j \le 2r}a_jx^j \in \mathbb R[x]$ is a nonnegative polynomial of odd degree $d=2r+1$, assumed to be unitary w.l.o.g. Then, in the limit as $x \longrightarrow -\infty$, we have the following inconsistent diagram
$$
0 \le \frac{f(x)}{x^{2r}} = x + o(1) \rightarrow -\infty,
$$
where the LHS is because $f(x) \ge 0$, and $x^{2r} > 0$ for $x \ne 0$.

Edit: correct solution for multivariate case
For the multivariate case, the story is similar. I've added a modification to address some counterexamples in the comments section. Apart from this small modification, the DNA of my strategy is essentially the same as before. We first prove the following lemma

Lemma. There always exist $s \in \mathbb R^n$ such that $m_s(x) := f(s_1 x,\ldots,s_n x) \in \mathbb R[x]$ is a unitary univariate real polynomial with the same degree $d$ as the $n$-variate polynomial $f(x_1,\ldots,x_n) \in \mathbb R[x_1,\ldots,x_n]$.

For example, in the counterexample of Robert Israel, simply take $(s_1,s_2) = (100, 1)$, to get $f(s_1 x, s_2 x) = 100 x^3 - x^3 + 100 x^2 = 99x^3 + 100 x^2$, a 3rd degree univariate polynomial.
Proof of Lemma.
To see why such a tuple exists, suppose w.l.o.g that $f$ only contains monomials of degree $d$ (everything else deleted), i.e suppose $f(x_1,\ldots,x_n) = \sum_{|\alpha| = d}c_\alpha \Pi_{i=1}^n x_i^{\alpha_i}$ where the $c_\alpha$'s a real numbers which are not all zero. Recall that for $\alpha_1,\ldots,\alpha_n \in \mathbb N$, the notation $|\alpha|$ is shorthand for $\alpha_1 + \dots + \alpha_n$.
Now, for $s \in \mathbb R^n$, consider the unitary univariate polynomial $m_s(x) := f(s_1x,\ldots, s_n x) \in \mathbb [x]$ of degree $\le d$. One computes
$$
m_s(x) = \sum_{|\alpha| = d}c_\alpha \Pi_{i=1}^n s_i^{\alpha_i}x_i^{\alpha_i} = \sum_{|\alpha| = d}c_\alpha x^{|\alpha|}\Pi_{i=1}^n s_i^{\alpha_i} = x^d\sum_{|\alpha| = d}c_\alpha \Pi_{i=1}^n s_i^{\alpha_i}=:x^d f(s_1,\ldots,s_n).
$$
But because $f$ is not an identically zero polynomial, we can always choose $s \in \mathbb R^n$ such that $f(s_1,\ldots,s_n) \ne 0$. For this choice of $s$, $m_s(x)$ has degree exactly equal to $d$, and we're done. $\Box$
Mindful of the above lemma, we can assume w.l.o.g that $f(x,\ldots,x)$ is a univariate polynomial of degree $d=2r+1$ (for there is always a subsitition $x_j \leftarrow s_j x_j$ for all $j=1,\ldots,n$, which lands us here). Thus $f(x,\ldots,x) = x^{2r+1} + \text{ lower order terms in }x$. Thus, in the limit as $x \longrightarrow -\infty$, we have
$$
0 \le \frac{f(x,\ldots,x)}{x^{2r}} = x + o(1)\rightarrow -\infty,
$$
which is again, an inconsistent diagram.

We conclude that any polynomial nonnegative polynomial in any number of variables must have even degree!
