$\frac{ab}{\sqrt{ab+3c}}=\frac{ab}{\sqrt{ab+c(a+b+c)}}=\frac{ab}{\sqrt{(c+a)(c+b)}}\leq \frac{1}{2}(\frac{ab}{c+a}+\frac{ab}{c+b})$tương tự:
$P\leq \frac{1}{2}(\frac{ab}{c+a}+\frac{ab}{c+b}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{b+a}+\frac{ac}{b+c})=\frac{1}{2}(a+b+c)\leq \frac{3}{2}$
dấu bằng xảy ra $\Leftrightarrow a=b=c=1$