Question

For three numbers $a,b,c$ product of the average of the numbers, $a^2,b^2,c^2$ and $\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ cannot be less than

• A. $1$
• B. $3$
• C. $9$
• D. none of these

Answer 1

The correct option is A $1$

$\frac{a^2+b^2+c^2}{3}\ge {\left(a^2{b}^2{c}^2\right)}^{\frac{1}{3}}[\because A.M\ge G.M.]$

and $\frac{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}{3}\ge {(\frac{1}{a^2}\times \frac{1}{b^2}\times \frac{1}{c^2})}^{\frac{1}{3}}$

On multiplying, we get

$\left(\frac{a^2+b^2+c^2}{3}\right).\left(\frac{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}{3}\right)\ge 1$

$\therefore$ product of the averages of $a^2,b^2,c^2$ and $\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ cannot be less than $1.$