A convexity inequality The function $\mathbb{R}^n\to \mathbb{R}$, $x\mapsto |x|^p$ (where $p>1$) is convex and thus the inequality
$$|y|^p-|x|^p\ge p(y-x)\cdot x |x|^{p-2}$$ is valid. In some lecture notes of Peter Lindqvist, it is remarked that this inequality can be strengthened to
$$|y|^p-|x|^p\ge p(y-x)\cdot x |x|^{p-2} + C(p) |y-x|^p$$ (of course $C(p)>0$) at least for $p>2$.
Does anyone know a proof of this inequality?

(I'm expanding my comment.)
This is a two-dimensional problem. One may assume $x=(1,0)$, $y=(1+\alpha, \beta)$ with $\sqrt{\alpha^2+\beta^2}=:r$. Then
$$|y|^p-|x|^p=((1+\alpha)^2 +\beta^2)^{p/2}-1\ ,\qquad p(y-x)\cdot x |x|^{p-2}=p\alpha\ .$$
It follows that we have to prove an inequality of the form
$$(1+2\alpha+r^2)^{p/2}\geq 1 + p\alpha + Cr^p \qquad\qquad (1)$$
for a suitable $C$, and we may assume $p\geq2$. 
Putting $r:=0$ in (1) the statement reads $(1+2\alpha)^{p/2}\geq 1 + p\alpha$, and this is true for $p\geq2$ by Bernoulli's inequality. Now the derivative  of the left side of (1) with respect to $r$ is
$${p\over 2}(1+2\alpha +r^2)^{p/2 -1}\ 2r\geq {p\over 2}r^{p-2}\  2r=pr^{p-1}\ ,$$
and the derivative of the right side of (1) with respect to $r$ is $Cp r^{p-1}$. So if $C=1$ the left side of (1) grows faster with $r$ than the right side. It follows that (1) is true with $C=1$.
