Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a differentiable function that satisfies the following for some $m>0$:
$$ \|\nabla f(\mathbf{x}) - \nabla f(\mathbf{y})\|\geq m\|\mathbf{x}-\mathbf{y}\|,\, \forall \mathbf{x},\mathbf{y} \in \mathbb{R}^n \label{1}\tag{1}. $$
Under what condition \eqref{1} implies the following inequality \eqref{2}
$$ \langle \nabla f(\mathbf{x}) - \nabla f(\mathbf{y}), y-x \rangle \geq m\|\mathbf{x}-\mathbf{y}\|^2,\, \forall \mathbf{x},\mathbf{y} \in \mathbb{R}^n \;?\label{2}\tag{2} $$
Please show how you would get \eqref{2} from \eqref{1} if that condition exists. The gradient of any function that satisfies \eqref{2} is strictly monotone. $m$-strongly convex functions satisfy the condition \eqref{2} always. However, I am looking for a case where one can get \eqref{2} only using \eqref{1}.
Note: for $f(x)=\frac{1}{2}\|Ax-b\|^2$ we can make it work. I am looking for another cases.
Suppose $f$ satisfies $(1)$ and is also convex and supercoercive.
Then the domain of the Fenchel conjugate $f^*$ is the entire space $\mathbb{R}^n$.
The condition $(1$) is now equivalent to $$ \|x^*-y^*\| \geq m\|\nabla f^*(x^*)-\nabla f^*(y)\|,\quad \forall x^*,y^* $$ Then the Baillon-Haddad Theorem guarantees that $(2)$ holds.