Question

Let equation $x^3+p{x}^2+qx-q=0$ where $p,q,\in R-\left\{0\right\}$ has 3 real roots $\alpha ,\beta ,\lambda$ in $H.P.$, then
Minimum value of $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$ is
$($You may use the inequality $a^2+b^2+c^2\ge ab+bc+ca$ for any $a,b,c\in R)$

• A. $\frac{1}{3}$
• B. $1$
• C. $\frac{4}{3}$
• D. $3$

Answer 1

The correct option is A $\frac{1}{3}$

Given $x^3+p{x}^2+qx-q=0$

Roots of equation: $\alpha ,\beta ,\lambda$ for HP

General cubic equation $a{x}^3+b{x}^2+cx+d$

$\alpha +\beta +\lambda =(-1{)}^1\frac{b}{a}=(-1)\frac{p}{1}=-p\alpha \beta +\beta \lambda +\alpha \lambda ={(-1)}^2\frac{c}{a}=q\alpha \beta \lambda ={(-1)}^3\frac{-q}{1}=q$

AM $\ge$ GM

$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}\frac{\gamma \beta +\alpha \gamma +\alpha \beta }{\alpha \beta \gamma }\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}\frac{q}{q}\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}{1}^2\ge 3^2{\left[\frac{1}{\alpha \beta \gamma }\right]}^{2/3}\left(1\right)\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{2/3}\left(2\right)$

DIviding $\left(2\right)$ by $\left(1\right)$

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\ge \frac{1}{3}$