Question

If $l,m,n$ be three positive roots of the equation $x^3-a{x}^2+bx+48=0$, then the minimum value of $\frac{1}{l}+\frac{2}{m}+\frac{3}{n}$ is

• A. $1$
• B. $2$
• C. $\frac{-3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is C $\frac{-3}{2}$

we Know that, $A.M.\ge G.M.$

$⟹\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

let $a=\frac{1}{l},b=\frac{2}{m},c=\frac{3}{n}$

Therefore,

$\frac{1}{3}(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge \sqrt[3]{3\left(\frac{1\times 2\times 3}{lmn}\right)}$

$(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge 3\times \sqrt[3]{3\left(\frac{1\times 2\times 3}{lmn}\right)}$

Given, the roots of the polynomial $x^3-a{x}^2+bx+48=0$ are $l,m,n$

Therefore, the product of the roots $lmn=-\left(\frac{48}{1}\right)=-48$

Substituting $lmn=-48$ in the above equation

$(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge 3\times \sqrt[3]{3\left(\frac{6}{-48}\right)}$

$(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge 3\times \sqrt[3]{3\left(\frac{1}{-8}\right)}$

$(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge 3\times \sqrt[3]{3(-\frac{1}{2}{)}^3}$

$(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge 3\times (-\frac{1}{2})$

$(\frac{1}{l}+\frac{2}{m}+\frac{3}{n})\ge (-\frac{3}{2})$

therefore, the minimum value is $-\frac{3}{2}$