Prove that the following definitions of a strongly convex function $f: \mathbb{R}^n \to \mathbb{R}$ are equivalent:
\begin{gather*} f(y) \ge f(x) + \nabla f(x)^T (y-x) + \frac{\mu}{2} {\lVert y-x \rVert}_2^2, \quad \forall x,y \in \mathbb{R}^n \tag{A} \\ \end{gather*}
\begin{gather*} (\nabla f(x) - \nabla f(y))^T (x-y) \ge \mu {\lVert y-x \rVert}_2^2, \quad \forall x,y \in \mathbb{R}^n \tag{B} \\ \end{gather*}
$(A) \Rightarrow (B)$ is simple. The first three steps are basic algebra. The fourth step is summing the previous two rows.
\begin{gather*} f(y) - f(x) - \nabla f(x)^T (y-x) \ge \frac{\mu}{2} {\lVert y-x \rVert}_2^2, \quad \forall x,y \in \mathbb{R}^n \\ f(x) - f(y) - \nabla f(y)^T (x-y) \ge \frac{\mu}{2} {\lVert y-x \rVert}_2^2, \quad \forall x,y \in \mathbb{R}^n \\ f(y) - f(x) + \nabla f(x)^T (x-y) \ge \frac{\mu}{2} {\lVert y-x \rVert}_2^2, \quad \forall x,y \in \mathbb{R}^n \\ (\nabla f(x) - \nabla f(y))^T (x-y) \ge \mu {\lVert y-x \rVert}_2^2, \quad \forall x,y \in \mathbb{R}^n \\ \end{gather*}
How would I prove $(B) \Rightarrow (A)$?
Notice the similarity between condition $(A)$ and the necessary and sufficient first-order condition for convexity and the similarity of condition $(B)$ and the monotone gradient condition for convexity. This is not a coincidence. If you let $g(x)=f(x)-\frac{\mu}{2}\|x\|^2$ it's easy to see that condition $(B)$ implies that $$(\nabla g(x) - \nabla g(y))^T (x-y) \geq 0 \quad \forall\, x,y\in \mathbb{R}^n$$ This is precisely the monotone gradient condition for convexity of $g(x)$. Since $g(x)$ is a convex differentiable function the first-order condition must hold, namely $$g(y) \geq g(x) + \nabla g(x)^T (y - x) \geq 0 \quad \forall\, x,y\in \mathbb{R}^n$$
If you expand the above inequality you will arrive at condition $(A)$. The same argument could be used to prove the other implication: $(A)$ implies that $g(x)$ is convex and so the monotone gradient condition for $g(x)$ must hold.