In http://www.mathlinks.ro/Forum/viewtopic.php?t=32586 , it was proven that $BM=\sqrt{s\left(s-b\right)}$, where $s= \frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, $BM^2=s\left(s-b\right)$. On the other hand, by the Heron formula, the area X of triangle ABC equals $X=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, and by the half-angle formulas, $\tan\frac{B}{2}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}$. Thus, $BM^2=s\left(s-b\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}: \sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}$ $=X: \tan\frac{B}{2}=X\cot\frac{B}{2}$, completing the proof. By the way, Orl, this is not an inequality... Darij