# A convex polynomial inequality.

Let $\phi(x)$ be a convex polynomial of degree $m$ at least two. Note that for $x,q \in \mathbb{R}$ $$\phi(x) + \phi(q) - 2\phi(\frac{x+q}{2}) = \sum_{l=1}^{m/2}\frac{\phi^{(2l)}(\frac{x+q}{2})}{2^{2l-1}(2l)!}|x-q|^{2l}$$ is strictly positive unless $x=q$, because the slopes of secant lines to $\phi$ are increasing.

I have proven using naive calculus-type estimates that there is some $C > 0$ such that $$\sum_{k=2}^{m}\frac{\bigl|\phi^{(k)}(\frac{x+q}{2})\bigr|}{k!}|x-q|^{k} \leq C \sum_{l=1}^{m/2}\frac{\phi^{(2l)}(\frac{x+q}{2})}{2^{2l-1}(2l)!}|x-q|^{2l}$$ uniformly in $x$ and $q$. I now need to show that $C \leq 2^{m}$ suffices.

But my approach of splitting $\mathbb{R} \times \mathbb{R}^{+}$ into various $(\frac{x+q}{2},|x-q|)$ regions and using either asymptotics or compactness no longer seems good enough. Has anyone seen such an estimate before, or does anyone know of general theory from convex analysis that makes this easier?

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Crossposted at MO: mathoverflow.net/questions/106696/… – Noah Stein Sep 10 '12 at 13:56