Question

If a, b, c $>0$, then prove that $\frac{3a}{b}+\frac{4b}{c}+\frac{16c}{3a}\ge 12$.

Answer 1

WE know that for a finite number of real terms

A.M$\ge$ G.M

Consider the term to be $\frac{3a}{b},\frac{4b}{c},\frac{!6c}{3a}$

A.M $=\frac{\frac{3a}{b}+\frac{4b}{c}+\frac{16c}{3a}}{3}$

G.M $=\sqrt[3]{3\frac{3a}{b}+\frac{4b}{c}+\frac{16c}{3a}}=4$

A.M $\ge$ G.M

$\frac{3a}{b}+\frac{4b}{c}+\frac{16c}{3a}\ge 12$