KöMaL 2024 C1820

Prove that if $a$, $b$, $c>0$ and $a+b+c=1$, then $a)$ ${\frac{1-a^2}{b+c}+\frac{1-b^2}{c+a}+\frac{1-c^2}{a+b }=4}$, $b)$ ${\frac{1-a^3}{b+c}+\frac{1-b^3}{c+a}+\frac{1-c^3}{a+b }\geq \frac{13}{3}}$.