$a^2-ab+b^2 \geq ab\Leftrightarrow \frac{1}{\sqrt{a^2-ab+b^2}} \leq \frac{1}{\sqrt{ab}}$ Tương tự :
$\frac{1}{\sqrt{b^2-bc+c^2}} \leq \frac{1}{\sqrt{bc}}$
$\frac{1}{\sqrt{c^2-ca+a^2}} \leq \frac{1}{\sqrt{ca}}$
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$\Rightarrow A \leq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}} \leq \frac{1}{2}(\frac{1}{a}+\frac{1}{b}) +\frac{1}{2}(\frac{1}{b}+\frac{1}{c})+\frac{1}{2}(\frac{1}{c}+\frac{1}{a})=3$
$\Rightarrow Max_A=3$ khi $a=b=c=1$