1. Min
P = $\sqrt{a+1} + \sqrt{b+1} \ge 0$
Dấu bằng xảy ra khi $a=b=-1$
2. Max
$2 = a^2 + b^2 \ge \frac{(a+b)^2}{2}$
$\Rightarrow a + b \leq 2$
$P = \sqrt{a+1} + \sqrt{b+1} \leqslant \sqrt{(1+1)(a+1+b+1)} \leqslant \sqrt{8} = 2\sqrt{2}$
Dấu bằng xảy ra khi $a=b=1$