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 $a=b=1$