Ta có: $\sqrt{(a+b+c)^3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3}$
$=\sqrt{[a^3+b^3+c^3+3(a+b)(b+c)(c+a)][\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3(a+b)(b+c)(c+a)}{a^2b^2c^2}]}$
$\geq \sqrt{(\sum a^3).(\sum \frac{1}{a^3})}+\frac{3(a+b)(b+c)(c+a)}{abc}$
Ta có: $\frac{(a+b)(b+c)(c+a)}{abc}=\frac{(a+b+c)(ab+bc+ca)}{abc}-1=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1$
Do đó: $3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\geq 3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}+6$
Vì : $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$
Từ đó có đpcm.