Ta có:$\sqrt{(a+b+c)^{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{3}}$=$\sqrt{\left[ {\Sigma a^{3}+3(a+b)(b+c)(c+a)} \right].\left[ {\Sigma \frac{1}{a^{3}}+\frac{3(a+b)(b+c)(c+a)}{(abc)^{2}}} \right]}$
$\geq \sqrt{(\Sigma a^{3})(\Sigma \frac{1}{a^{3}}}+\frac{3(a+b)(b+c)(c+a)}{abc}$
Mà$\frac{(a+b)(b+c)(c+a)}{abc}=\frac{(a+b+c)(ab+bc+ca)}{abc}-1=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$
Do đó:$3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\geq 3(\Sigma a)(\Sigma \frac{1}{a})-3\sqrt{(\Sigma a)(\Sigma \frac{1}{a})}+6$
Lại có:$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$
$\Rightarrow (\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}-1)^{3}\geq 5+\sqrt{(\Sigma a^{3})(\Sigma \frac{1}{a^{3}})}$
$\Rightarrow đpcm$
Dấu''='' xra$\Leftrightarrow a=b=c$