If a,b,c are three positive numbers then using cauchy schwerz inequality prove that

$(a+b+c)(\frac1a+\frac1b+\frac1c) \geq 3^2$

Consider two vectors

$\begin{align} u &= ( \sqrt a , \sqrt b, \sqrt c )\\ v &= (\frac{1}{\sqrt a} , \frac{1}{\sqrt b}, \frac{1}{\sqrt c})\\ u.v &= ( \sqrt a , \sqrt b, \sqrt c ) (\frac{1}{\sqrt a} , \frac{1}{\sqrt b}, \frac{1}{\sqrt c})\\ u.v &= (\frac{ \sqrt a}{\sqrt a} +\frac{\sqrt b}{\sqrt b}+ \frac{\sqrt c }{\sqrt c})\\ u.v &= (1+1+1) \\ u.v &= 3 \\ \\ \lvert \lvert u \lvert \lvert & = \sqrt{\sqrt a ^2 + \sqrt b ^2 + \sqrt c ^2 } \\ \lvert \lvert u \lvert \lvert & = \sqrt{a+b+c} \\ {\lvert \lvert u \lvert \lvert } ^2 & = {a+b+c} \\ \\ \lvert \lvert v \lvert \lvert & = \sqrt{[\frac{1}{\sqrt a}]^2+[\frac{1}{\sqrt b}]^2+[\frac{1}{\sqrt c}]^2}\\ \lvert \lvert v \lvert \lvert & = \sqrt{\frac1a+\frac1b+\frac1c}\\ {\lvert \lvert v \lvert \lvert }^2 & = {\frac1a+\frac1b+\frac1c}\\ \end{align}$

by cauchy-schwartz inquality

$(u.v)^2 \leq {\lvert \lvert u \lvert \lvert } ^2 . {\lvert \lvert v \lvert \lvert }^2 $

$ (3)^2 \leq ({a+b+c}) ({\frac1a+\frac1b+\frac1c}) - - - -$ Hence Proved.