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Q.
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
Statistics
Solution:
$\frac{a^{2}+b^{2}+c^{2}}{3} \ge\, \left(a^{2}\,b^{2}\,c^{2}\right)^{1 /3}$ using A.M. $\ge$ GM
and $\frac{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}{3} \ge \left(\frac{1}{a^{2}} \frac{1}{b^{2}} \frac{1}{c^{2}}\right)^{1/ 3}$ on multiplying
$\therefore \left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)\cdot\left(\frac{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}{3}\right)\ge\,1$
$\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 $