We have a twice continuously differentiable function $f:\mathbb{R}^d\to\mathbb{R}$ and $\mu,L>0$ constants
We have $\mu\cdot E\preccurlyeq\nabla^2f(x)\preccurlyeq L\cdot E$. that means that the eigenvalues are bounded above and away from zero (E is the identity matrix)
We have that $f$ is convex
Now we want to show:
$(\nabla f(x)-\nabla f(y))^T\cdot(x-y)\geq\frac{1}{L}||\nabla f(x)-\nabla f(y)||^2$
This is called the co-coercivity of the gradient.
The following is inspired/stolen from these wonderful notes of Vandenberghe.
Suppose $g$ is convex, $L$ Lipschitz smooth and has a minimiser $y^*$. Note that for any $x$ $g(y) \le g(x) + \langle \nabla g(x), y-x \rangle +{L \over 2} \|y-x\|^2$, and minimising both sides gives $g(y^*) \le g(x)-{1 \over 2L} \| \nabla g(x) \|^2$ (a lower bound on the 'cost to go').
Now pick some $y$ and define $\phi(x) = f(x)-\langle \nabla f(y),x \rangle $, note that $\phi$ is convex, $L$ Lipschitz smooth and has a minimiser at $x=y$ (since $\nabla \phi(y) = 0$). Hence $\phi(x)-\phi(y) \ge {1 \over 2L} \| \nabla \phi(x) \|^2$ and unraveling gives $f(x)-f(y) - \langle \nabla f(y), x-y \rangle \ge {1 \over 2L} \| \nabla f(x) -\nabla f (y)\|^2 $.
Exchanging the roles of $x,y$ gives $f(y)-f(x) - \langle \nabla f(x), y-x \rangle \ge {1 \over 2L} \| \nabla f(x) -\nabla f (y)\|^2 $ and adding yields $\langle \nabla f(x) - \nabla f(y), x-y \rangle \ge {1 \over L} \| \nabla f(x) -\nabla f (y)\|^2 $