thelazyboy

V.$(\sqrt{x-1}+\sqrt{y+2})^{2}\leq 2(x+y+1)
\Rightarrow (x+y+2)^{2}\leq 8(x+y+1)
Đặt t=x+y+1
\Rightarrow 3-2\sqrt{2}\leq t\leq 3+2\sqrt{2}$
Q$\geqslant \frac{4}{x+y+3}\geq \frac{4}{t+2}\geq \frac{4}{5-2\sqrt{2}}$