# 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.

• Title should say "always nonnegative" not "always positive" Commented Apr 8, 2013 at 22:15
• @Hammerite Edited, thanks. Commented Nov 22, 2013 at 16:59

Move $-1008x^2$ to the right side and use $\mathrm{AM}\geq\mathrm{GM}$ $$\frac{x^{2016}+\underbrace{1+1+\ldots+1}_{1007}}{1008} \geq \sqrt[1008]{x^{2016} \underbrace{1 \cdot 1\cdots 1}_{1007}} =x^2$$

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.

Any real polynomial that is $\ge 0$ on the real line is a sum of squares of real polynomials. But just because it is true, doesn't mean it can be done in an elegant way.

Suppose we are proving this by induction on the degree, and we have already done up to degree 2015. Then we come to this polynomial $P(x) = x^{2016}-1008x^2+1007$. As Abel showed, polynomial $P$ has minimum value $0$ at $x=1$. Because it is a minimum, in fact $x=1$ is an (at least) double root. So $P(x) = (x-1)^2Q(x)$. Now (by our induction hypothesis) since $Q$ is nonnegative everywhere, it is a sum of squares, so multiplying that decomposition by $(x-1)^2$ we get a decomposition for $P$ as a sum of squares.

The other type of step in the inductive proof would be a case where the polynomial has positive minimum. Then it is a positive (hence square) constant plus a polynomial with minimum zero, and then proceed as above.