This can be directly computed: If $M$ is the midpoint of $BC$, then we have two right triangles $ABE$ and $BEM$ with sides $AB=a, BE=b, AE=c=\sqrt{a^2+b^2}$ and $EM=h=\frac{ab}{c}$ so that $BM=\sqrt{b^2-h^2}=\frac{b^2}{c}$. So the diameter is $2b$ and the length of the medial line is $c+\frac{b^2}{c}$ and we are done by AM-GM (note that $b \ne c$ so equality is not possible).