$\big\langle\nabla f(x)-\nabla f(y),\,(x-y)\big\rangle\ge m\,\left\| x-y\right\|^2,\;m>0\;$ for strictly and strong convex function 
Prove that $\,\big\langle\nabla f(x) - \nabla f(y),\, (x-y)\big\rangle \geq  m\,\left\lVert x-y\right\rVert^2, \;\,m > 0\,$ for strong convex function $f: \mathbb R^n \to\mathbb R$. Is it true for strictly convex function.

I have to clue to prove above statement. Can anyone give me some hints?
 A: We wish to show that if a differentiable function $f:\mathbb R^n \to \mathbb R$ is strongly convex with parameter $m > 0$ then it is strongly monotone with parameter $m$.
To say that $f$ is strongly convex with parameter $m$ means that the function
$h(x) = f(x) - \frac{m}{2} \|x\|^2$ is convex.  (At least, that's my favorite definition of strong convexity.  It has the advantage of making sense even for nondifferentiable functions.)
Because $h$ is a convex function, it is monotone:
\begin{align}
&\langle \nabla h(x) - \nabla h(y),  x - y \rangle \geq 0 \quad \text{for all } x,y \\
\implies & \langle \nabla f(x)- mx - (\nabla f(y) - my), x - y \rangle \geq 0 \quad \text{for all } x,y \\
\implies & \langle \nabla f(x) - \nabla f(y), x - y \rangle \geq m \| x - y \|^2 \quad \text{for all } x,y.
\end{align}
Thus $f$ is strongly monotone with parameter $m$.
A: The function $f(x) = x^4$ is strictly convex and the above inequality does not hold for any $m>0$.
In particular, if $m>0$ and
if $y=0$, the left hand side of the above formula is 
$(f'(x)-f'(0))(x-0) = 4x^4$, and for $0<|x| < {\sqrt{m} \over 2}$, we have
$f'(x)-f'(0))(x-0) < m(x-0)^2$.
