Sequences And Series Ques 26
- If $a>0, b>0$ and $c>0$, then prove that
$ (a+b+c) \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9 $
Show Answer
Solution: Since $a, b, c>0$
$\Rightarrow \quad \frac{(a+b+c)}{3}>(a b c)^{1 / 3}$ $\quad$ ……..(i)
[using AM $\geq$ GM]
Also, $\quad \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq\left(\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}\right)^{1 / 3}$ $\quad$ ……..(ii)
[using AM $\geq$ GM ]
On multiplying Eqs. (i) and (ii), we get
$ \begin{aligned} & \frac{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{9} \geq(a b c)^{1 / 3} \frac{1}{(a b c)^{1 / 3}} \\ & (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9 \end{aligned} $
BETA
