Ex. 72. The height of a right triangle, dropped to the hypotenuse, is $h$. What is the minimum length that the median, which bisects the larger leg, can have?
Ex. 72. $\frac{3}{2} h$. Solution. Let the segments into which the height $h$ divides the hypotenuse be denoted by $x$ and $y (x \geq y)$. Then $m^{2}=\left(\frac{x}{2}+y\right)^{2}+\left(\frac{h}{2}\right)^{2}=\frac{x^{2}}{4}+x y+y^{2}+\frac{h^{2}}{4}=\frac{5 h^{2}}{4}+\frac{1}{4}\left(x^{2}+\frac{4 h^{4}}{x^{2}}\right)$. By the inequality between the arithmetic mean and the geometric mean, $x^{2}+\frac{4 h^{4}}{x^{2}} \geq 2 \sqrt{4 h^{4}}=4 h^{2}$. From this, we obtain that $m^{2} \geq \frac{5 h^{2}}{4}+h^{2}=\frac{9}{4} h^{2} \Rightarrow m \geq \frac{3}{2} h$.