Polynomial always non-negative Is there an elegant way to demonstrate that (for example) $x^{2016}-1008x^2+1007\ge 0$ $\forall x\in \mathbb{R}$ ? I tried to write it as sum of squares, but I didn't succeed.

Alternatively, an analytic approach:
Lets call your polynomial $P(x) = 2^{2016}-1008x^2+1007$.
Since $\lim_{x\to\pm\infty} P(x) = \infty$, $P$ has a global minimum.
To find this global minimum we find the roots of $P^\prime$.
$P^\prime(x) = 2016x^{2015}-2016x = 2016x(x^{2014}-1)$, the roots of which are $x=0$ and $x=\pm1$.
Evaluating $P(0)$ gives $1007$ and $P(\pm1)=0$, thus the global minimum of $P$ is $0$, which is exactly what you wanted to show.
