There is a $\sigma$-strongly convex function, $f(x')\ge f(x)+ \langle x'-x,\mu\rangle +\frac{\sigma}{2}\left|x'-x\right|^2$ where $\mu \in \partial f(x)$, $\mu ' \in \partial f(x')$.
How could I get the follow equation, $f(x')\le f(x)+\langle x'-x,\mu\rangle+\frac{1}{2\sigma}\left|\mu '-\mu\right|^2$
This is equivalent to show,
$$\frac{1}{2\sigma}\left|\mu '-\mu\right|^2\ge \frac{\sigma}{2}\left|x'-x\right|^2$$ $$\frac{1}{\sigma ^2}\left|\mu '-\mu\right|^2\ge \left|x'-x\right|^2$$
I assume that your first inequality holds also for $x$ and $x'$ swapped, i.e., $$f(x) \ge f(x') + \langle x - x', \mu' \rangle + \frac\sigma{2} | x - x' |^2.$$ Adding this to your first inequality, we obtain
$$0 \ge \langle x - x', \mu' - \mu \rangle + \sigma \,| x - x' |^2.$$ Using Cauchy-Schwarz, we have $$ \langle x' - x, \mu - \mu' \rangle \le |x-x'||\mu-\mu'| $$ $$0 \ge - |x - x'| \, |\mu - \mu'| + \sigma|x-x'|^2.$$ Rearranging terms yields the desired inequality $$|\mu - \mu'| \ge \sigma\,|x-x'|.$$