Popoviciu's inequality on variances

Let ${\displaystyle A}$ be a random variable with mean ${\displaystyle \mu }$, variance ${\displaystyle \sigma ^{2}}$, and ${\displaystyle \Pr(m\leq A\leq M)=1}$. Then, since ${\displaystyle m\leq A\leq M}$,

${\displaystyle 0\leq \mathbb {E} [(M-A)(A-m)]=-\mathbb {E} [A^{2}]-mM+(m+M)\mu }$.

Thus,

${\displaystyle \sigma ^{2}=\mathbb {E} [A^{2}]-\mu ^{2}\leq -mM+(m+M)\mu -\mu ^{2}=(M-\mu )(\mu -m)}$.

Now, applying the Inequality of arithmetic and geometric means, ${\displaystyle ab\leq \left({\frac {a+b}{2}}\right)^{2}}$, with ${\displaystyle a=M-\mu }$ and ${\displaystyle b=\mu -m}$, yields the desired result:

${\displaystyle \sigma ^{2}\leq (M-\mu )(\mu -m)\leq {\frac {\left(M-m\right)^{2}}{4}}}$.