Ta có $P=\sum_{}^{cyc} \frac{ab}{\sqrt{2c+ab}}=\sum_{}^{cyc} \frac{ab}{\sqrt{c(a+b+c)+ab}}$ $=\sum_{}^{cyc}\frac{ab}{\sqrt{(a+c)(b+c)}}\leq \frac{1}{2}(\sum_{}^{cyc}\frac{ab}{a+c}+ \sum_{}^{cyc}\frac{ab}{b+c})=\frac{1}{2}(a+b+c)=1 $
Dấu đẳng thức $a=b=c=\frac{2}{3}$
Vậy $MaxP=1$