P=$\frac{\sqrt{a}}{\sqrt{b}+\sqrt{a+b}}+ \frac{\sqrt{b}}{\sqrt{a}+\sqrt{a+b}}$
Ta có: $\sqrt{a}+\sqrt{b}\geqslant \sqrt{a+b}$
$\Rightarrow \sqrt{b} + \sqrt{a+b}\leqslant \sqrt{a}+\sqrt{a}+\sqrt{b}=2\sqrt{a}+\sqrt{b}$
$\Rightarrow \frac{\sqrt{a}}{\sqrt{b}+\sqrt{a+b}} \geqslant \frac{\sqrt{a}}{2\sqrt{b}+\sqrt{a}}$
Tương tự, ta cũng có:
$\frac{\sqrt{b}}{\sqrt{a}+\sqrt{a+b}} \geq \frac{\sqrt{b}}{2\sqrt{a}+\sqrt{b}}$
$\Rightarrow P \geq \frac{\sqrt{a}}{2\sqrt{b}+\sqrt{a}} + \frac{\sqrt{b}}{2\sqrt{a}+\sqrt{b}}$
$\Leftrightarrow P \geq 2\frac{a+b}{2\sqrt{ab}}$
Mà $a+b\geq 2\sqrt{ab}$
$\Rightarrow P \geq 1$