Lower bound for tail probability

The function $f(t):= \mathbb P(|X-\mu|\ge t)-\frac t{2M}$ is strictly decreasing on $[0,\infty)$ and we have $f(0)=1$, $f(2M)=-1$. Let $s=\inf\{t\ge0\mid f(t)\le 0\}$. Then $$ \sigma^2 = \int (x-\mu)^2 d\mu =\int_0^\infty 2r\mathbb P(|X-\mu|\ge r) dr\\ \le\int_0^s2r\cdot1\,dr+\int_s^{2M}2r\cdot\frac s{2M}dr\\ =s^2+2Ms-\frac{s^3}{2M}. $$ If $s=0$, then this shows $\sigma^2=0$ and thus the claim we want to show is the trivial statement $\mathbb P(|X-\mu|\ge0)\ge0$. Therefore, we may assume $0< s\le 2M$ and conclude $$ \sigma^2\le s^2+2Ms-\frac{s^3}{2M}< 2Ms+2Ms-0=4Ms.$$ But then $\frac{\sigma^2}{4M}<s$ implies $f\left(\frac{\sigma^2}{4M}\right)>0$, i.e. $$ \mathbb P\left(|X-\mu|\ge \frac{\sigma^2}{4M}\right)>\frac{\sigma^2}{8M^2}.$$ Apparently, the only case where $\le$ canot be replaced with $>$ is the case of an (almost surely) constant variable.