$a+2b=1\\ \Leftrightarrow a=1-2b\\ P=a^2+2b^2\\ =(1-2b)^2+2b^2\\ =6b^2−4b+1\\ =\left(\sqrt{6}b\right)^2-2.\sqrt{6}b.\dfrac{\sqrt{6}}{3}+\left(\dfrac{\sqrt{6}}{3}\right)^2+\dfrac{1}{3}\\ =\left(\sqrt{6}b-\dfrac{\sqrt{6}}{3}\right)^2+\dfrac{1}{3} \ge \dfrac{1}{3} \, \forall \, b$
Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} a=1-2b\\\sqrt{6}b-\dfrac{\sqrt{6}}{3}=0\end{array} \right.$
$\Leftrightarrow \left\{\begin{array}{l} a=\dfrac{1}{3}\\b=\dfrac{1}{3}\end{array} \right.$
Vậy $max_P=\dfrac{1}{3}$ xảy ra khi $a=b=\dfrac{1}{3}$
