Problem

Source: Moldova TST 2018

Tags: inequalities



The positive real numbers $a,b, c,d$ satisfy the equality $ \frac {1}{a+1} + \frac {1}{b+1} + \frac {1}{c+1} + \frac{ 1}{d+1} = 3 $ . Prove the inequality $\sqrt [3]{abc} + \sqrt [3]{bcd} + \sqrt [3]{cda} + \sqrt [3]{dab} \le \frac {4}{3} $.