I am having trouble with the following proof:
Prove that for every three positive real numbers a, b, and c that
$(a+b+c)*(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \ge 9$.
I have tried to directly prove this but all I get are dead ends.
I am having trouble with the following proof:
Prove that for every three positive real numbers a, b, and c that
$(a+b+c)*(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \ge 9$.
I have tried to directly prove this but all I get are dead ends.
This is an answer by a different method of those given by the link of André Nicolas.
By Cauchy-Schwarz inequality we have $$\left(\sum_i a_i b_i\right)^2\le \left(\sum_i a_i^2\right)\left(\sum_i b_i^2\right)$$ so it suffices to take $a_i=\sqrt a$ and $b_i=b^{-1/2}$ and we have the desired result.