Question 24
(THA 2) Let $a,b,c,d$ be nonnegative real numbers such that $ab+bc+cd+da=1$. Show that
$$\frac{a^3}{b+c+d}+\frac{b^3}{a+c+d}+\frac{c^3}{a+b+d}+\frac{d^3}{a+b+c}\ge\frac13.$$
$$\frac{a^3}{b+c+d}+\frac{b^3}{a+c+d}+\frac{c^3}{a+b+d}+\frac{d^3}{a+b+c}\ge\frac13.$$