MathDB
Another nice Inequality

Source: Bulgaria 2016,P 3

July 16, 2016
inequalitiesalgebrainequalities proposed

Problem Statement

For a,b,c,d>0a,b,c,d>0 prove that a+ab+abc3+abcd44a.a+b2.a+b+c3.a+b+c+d44\frac {a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4} \leq \sqrt[4]{a.\frac{a+b}{2}.\frac{a+b+c}{3}.\frac{a+b+c+d}{4}}