Question

Let equation $x^3+p{x}^2+qx-q=0$ where $p,q\in R-\{0\}$ has 3 real roots $\alpha ,\beta ,\gamma$ 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

Answer: (A)

$2\left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\right)\ge 2\left(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }\right)$
using $(a^2+b^2+c^2\ge ab+bc+ca$ with $a=\frac{1}{\alpha },b=\frac{1}{\beta }$ and $c=\frac{1}{\gamma })$
$\Rightarrow 3\left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\right)\ge {\left(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\right)}^2=1$ (add $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$ both sides)
$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\ge \frac{1}{3}$