4. In a right triangle $A B C$ with $\angle C=90^{\circ}, B C=10$, and $A C=15$. Two squares are inscribed in $A B C$ as shown in the figure. Find the minimum sum of the areas of the squares.
Answer: 36
Solution: Let $N M=s$ and $S R=t$ be the side lengths of the two squares. By the Pythagorean theorem, we have $A B=5 \sqrt{13}$. From the above figure, triangles $N M B, A C B$ and $A R S$ are similar. Thus, $M B=\frac{B C \cdot N M}{A C}=\frac{2 s}{3}$ and $A R=\frac{A C \cdot R S}{B C}=\frac{3 t}{2}$. We see that $5 \sqrt{13}=A B=A R+R Q+Q M+$ $M B=\frac{5 s}{3}+\frac{5 t}{2}$ or $\sqrt{13}=\frac{s}{3}+\frac{t}{2}$. Hence, by Cauchy-Schwarz inequality, we get
$$
\left(s^{2}+t^{2}\right)\left(\frac{1}{3^{2}}+\frac{1}{2^{2}}\right) \geq\left(\frac{s}{3}+\frac{t}{2}\right)^{2}=13
$$
so that $s^{2}+t^{2} \geq 13 /\left(\frac{1}{9}+\frac{1}{4}\right)=36$. Hence, the minimum sum of the areas of the squares $S_{1}$ and $S_{2}$ is 36 , which is attained when $s=12 / \sqrt{13}$ and $t=18 / \sqrt{13}$.