Inequality with 4 variables, powers.

Question

$\{ a,b,c,d \in\Bbb R_+\ \}$
$a^6 + b^3 + c^2 + d \ge 2 \sqrt[3]{2} \sqrt[4]{3} \sqrt {abcd} $

I have no idea how to do this sort of inequalities, never seen it before.

Answer 1

By the AM-GM inequality, one has $$ a^6+b^3+c^2+d=a^6+\frac{1}{2}b^3+\frac{1}{2}b^3+\frac{1}{3}c^2+\frac{1}{3}c^2+\frac{1}{3}c^2+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d\ge 12\sqrt[12]{a^6(\frac12b^3)^2(\frac13c^2)^3(\frac16d)^6}$$ After simple calculation you will get the answer.