Cách 2:
Ta có: $\frac{a}{a+ab}+\frac{b}{b+bc}=\frac{a}{(a+b)(a+c)}+\frac{b}{(a+b)(b+c)}$$=\frac{(ab)-(x^2+b^2)}{(a+b)(1-a)(1-b)}=$$\frac{(a+b)-(a+b)^2+2ab}{(a+b)[1-(a+b)+ab]}$$\leq\frac{(a+b)-(a+b)^2+\frac{(a+b)^2}{4}}{(a+b)[1-(a+b)+\frac{(a+b)^2}{4}]}=\frac{2}{c+1}$
$\rightarrow \frac{\sqrt{abc}}{c+ab}\leq \frac{\sqrt{c}(a+b)}{2[c+\frac{(a+b)^2}{a}]}=\frac{2\sqrt{c}(1-c)}{(c+1)^2}$
Khi đó:
$P\leq \frac{2}{c+1}+\frac{2\sqrt{c}(1-c)}{(c+1)^2}$
Đặt: $t=\sqrt{c}\rightarrow 0<t<1.$
Xét hàm số:
$f(t0=\frac{2(-t^3+t^2+t+1)}{(t^2+1)^2}$
Có:
$f'(t)=\frac{2(t^4-2t^3-6t^2-2t+1)}{(t^2+1)^3}$
và $f'(t)=0$
$\Leftrightarrow ..............\Leftrightarrow t=2-\sqrt{3}$
$\rightarrow ................$