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Q.
If a, b, c are any three positive numbers, then the least value of $\left(a+b+c\right) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) $ is
Statistics
Solution:
A.M. of any number of non-negative numbers is always greater than or equal to the H.M. between them, therefore,
$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a} + \frac{1}{b}+\frac{1}{c}} $
$\Rightarrow \left(a+b+c\right) \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge 9 $