Monotonicity inequality How can we prove this monotonicity inequality 
let $1<p<\infty$ we have the vector valued fuction $a:\mathbb{R}^N\rightarrow\mathbb{R}^N$defined by $ a(x)=|\xi|^{p-2}\xi$ for $\xi\neq0$, $a(0)=0$
.If $1<p<\infty$,then we have 
$(a(\xi)-a(\xi')).(\xi-\xi')>0 \quad \forall \xi,\xi'\in\mathbb{R}^N,\xi\neq\xi'$
.If $2\leq p<\infty$,then there exists a constant $c>0$ such that 
$(a(\xi)-a(\xi')).(\xi-\xi)\geq c|\xi-\xi'|^p \quad \forall \xi,\xi'\in\mathbb{R}^N$
 A: Let $f:x\mapsto \|x\|^p$. Since $x\mapsto \|x\|$ is convex and $a\mapsto a^p$ is convex and increasing over $\mathbb R^+$, $f$ is convex. Using the expression $f(x) = (\|x\|^2)^{p/2}$ the gradient of $f$ can be computed using the chain rule as $\|x\|^{p-2}x$. 
Writing the inequality on gradients of convex functions proves the claim.

Expanding $ \|x+h\|^p$ as $ \displaystyle \|x\|^p\left(1+\frac{2x^Th}{\|x\|^2}+\frac{\|h\|^2}{\|x\|^2} \right)^{p/2}$ and writing the Taylor-Young expansion of $(1+x)^{p/2}$ shows that the Hessian of $f$ at $x$ is $\displaystyle\frac{p(p-2)}{2}\|x\|^{p-4}xx^T$, the eigenvalues of which are $\geq 0$ (because $p\geq 2$).
For $m$ sufficiently large, the eigenvalues of $\displaystyle\frac{p(p-2)}{2}\|x\|^{p-4}xx^T-mI_N$ are $\leq 0$, hence $x\mapsto f(x)-\frac m2 \|x\|^2$ is concave. A proof similar to that encountered when dealing with strongly convex functions shows that $\langle x-y, \nabla f(x)-\nabla f(y)\rangle \leq m \|x-y\|^2$ which resembles the inequality wanted (except for the square in the RHS).
