Question 34
Let $a$, $b$, $c$, $d$ be non-negative real numbers such that
$$
\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} = 3.
$$
Prove that
$$
3(ab + ac + ad + bc + bd + cd) + \frac{4}{a+b+c+d} \le 5.
$$
$$
\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} = 3.
$$
Prove that
$$
3(ab + ac + ad + bc + bd + cd) + \frac{4}{a+b+c+d} \le 5.
$$