Cách 2:
$A=\Sigma \frac{\frac{a}{b}}{\sqrt{\frac{a}{b}+1}}\geq \frac{(\sqrt{\frac{a}{b}})^2}{\Sigma \sqrt{\frac{a}{b}+1}}$ $(1)$Đặt $(x;y;z)=(\frac{a}{b};\frac{b}{c};\frac{c}{a})\rightarrow xyz=1$
Khi đó: $(1)\Leftrightarrow \frac{(\Sigma \sqrt{x})^2}{\Sigma \sqrt{x+1}}\geq \frac{\Sigma x+2\Sigma \sqrt{xy}}{\sqrt{3(\Sigma x+3)}}\geq \frac{\Sigma x+6}{..........}$
So that: $A\geq \frac{S+3}{\sqrt{3S}}(S=\Sigma x+3\geq 6)$
Mà: $\sqrt{S}+\frac{3}{\sqrt{S}=\frac{\sqrt{S}}{2}}+(\frac{\sqrt{S}}{2}+\frac{3}{\sqrt{S}})\geq \frac{\sqrt{6}}{2}+\frac{2\sqrt{3}}{2}=\frac{2\sqrt{3}}{2}$
$\Leftrightarrow \frac{S+3}{\sqrt{3S}}\geq \frac{3\sqrt{2}}{2}$
Đẳng thức khi $a=b=c./$