Problem

Source: 2018 Taiwan TST Round 1

Tags: inequalities, Taiwan



Let $ a,b,c,d $ be four non-negative reals such that $ a+b+c+d = 4 $. Prove that $$ a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b} \ge 4\sqrt{5} $$