$f$ is $\alpha$-strongly convex if there exist a constant $\alpha$ such that $$ f(y) \geq f(x)+\left<f '(x),y-x\right>+\frac{\alpha}{2}\|y-x\|^2 \quad (1)$$ or $$ \left<f'(y)-f '(x),y-x\right> \geq \alpha\|y-x\|^2 \quad (2)$$
for all $x,y$.
I know how to get the secone formula from the first one but I don't know how to get the first one from the second. That is: how to prove $(1)$ from $(2)$. I find the most related one is this question, but it only talks about how to prove $f$ is convex, not $f$ is strongly convex.
$$\begin{split} f(y)-f(x) - f'(x)(y-x)& = \int_0^1 (f'(x+s(y-x))-f'(x))(y-x) ds\\ &= \int_0^1 s^{-1} (f'(x+s(y-x))-f'(x))(s(y-x) ) ds\\ &\ge \alpha \int_0^1 s^{-1}s^2\|y-x\|^2 ds= \frac\alpha2 \|y-x\|^2 \end{split}$$