Question 32
Prove for all positive real numbers $a$, $b$, $c$, $d$ that
$$
\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} \ge \frac{4a+4b+4c-3d}{9}.
$$
$$
\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} \ge \frac{4a+4b+4c-3d}{9}.
$$