1. Áp dụng BĐT Bunhia ta có
$a^2+2b^2=a^2+b^2+b^2 \ge \frac13(a+b+b)^2=\frac13(a+2b)^2$$\Rightarrow \sqrt{a^2+2b^2} \ge \frac1{\sqrt3}(a+2b)$.
Tương tự ta có
$\sqrt{b^2+2a^2} \ge \frac1{\sqrt3}(2a+b)$.
Suy ra
$\sqrt{a^2+2b^2}+\sqrt{b^2+2a^2} \ge \frac1{\sqrt3}(3a+3b)=\sqrt3(a+b)$
$\Rightarrow P= \frac{a+b}{\sqrt{a^2+2b^2}+\sqrt{b^2+2a^2}} \le \sqrt 3$.
Vậy $\max P=\sqrt 3\Leftrightarrow a=b.$