Question

Let equation $x ^3+ px ^2+ qx - q =0$ 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 $a^2+b^2+c^2 \geq a b+b c+c a$ for any a, b, c $\in R$ )

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

Answer 1

Answer: (A)

$2\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right) \geq 2\left(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}\right)$
using $\left(a^2+b^2+c^2 \geq a b+b c+c a\right.$ with $a=\frac{1}{\alpha}, b=\frac{1}{\beta}$ and $\left.c=\frac{1}{\gamma}\right)$
$\Rightarrow 3\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right) \geq\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)
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} \geq \frac{1}{3}$