$4P^{2}\leq 3.(\frac{4a}{a+1}+\frac{4b}{b+1}+\frac{4c}{c+1})$
<=> $4P^{2}\leq 3.(\frac{4a}{a+b+a+c}+\frac{4b}{a+b+b+c}+\frac{4c}{a+c+b+c})$
<=> $4P^{2}\leq 3.(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c})$
<=> $4P^{2}\leq 3.3$
<=> $P\leq \frac{3}{2}$
Vậy P max = $\frac{3}{2}$ <=> a=b=c=$\frac{1}{3}$