Question

If $a,b,c$ are three positive roots of the equation $x^3-p{x}^2+qx-48=0$ $(p,q>0)$, then find the minimum value of $2(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})$.

Answer 1

$2(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})$

$\because AM\ge GM$

$=>\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\ge 0$

$=>\frac{\frac{1}{a}+\frac{2}{b}+\frac{3}{c}}{3}\ge \sqrt[\frac{1}{3}]{\frac{1}{3}\frac{1}{a}\times \frac{2}{b}\times \frac{3}{c}}$

$=>\frac{1}{a}+\frac{2}{b}\times \frac{3}{c}\ge 3{\left(\frac{6}{abc}\right)}^{\frac{1}{3}}$

$\ge 3{\left(\frac{6}{48}\right)}^{\frac{1}{3}}$

$\ge 3{\left(\frac{1}{2}\right)}^{3\times \frac{1}{3}}$

$=>\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\ge \frac{3}{2}$

$=>2(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})\ge 3$

Hence minimum will be $3$.