Ta có $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) $
= $\frac{a}{a} + \frac{b}{b} + \frac{c}{c}$ + $\frac{a}{b}$ +$\frac{a}{c}$ +$\frac{b}{a}$ +$\frac{b}{c}$
+$\frac{c}{b}$ +$\frac{c}{a}$
= $1 + 1 + 1 +$ $+$ ($\frac{a}{b}$ + $\frac{b}{a}$)$ + $ ($\frac{a}{c}$$+$$\frac{c}{a}$ ) $+$ ($\frac{b}{c}$+ $\frac{c} {b}$)
= $3 + $ ($\frac{a}{b}$ + $\frac{b}{a}$) + ($\frac{a}{c}$+$\frac{c}{a}$ ) + ($\frac{b}{c}$+$\frac{c} {b}$)
Vì $a,b,c >0$ nên áp dụng bđt Cô si ta có
$\frac{a}{b}$ + $\frac{b}{a}$ $\geq 2\sqrt[]{\frac{a}{b}.\frac{b}{a}}$ = 2
$\frac{a}{c}$+$\frac{c}{a}$ $\geq 2\sqrt[]{\frac{a}{c}.\frac{c}{a}}$ = 2
$\frac{b}{c}$+$\frac{c}{b}$ $\geq 2\sqrt[]{\frac{b}{c}.\frac{c}{b}}$ = 2
=> $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) $ $\geq$ 3 + 2 + 2 +2 = 9
=> đpcm
Dấu "=" xảy ra khi a = b = c
