Ta có:$P=a+\frac{1}{b(a-b)^{2}}=\frac{a-b}{2}+\frac{a-b}{2}+b+\frac{1}{b(a-b)^{2}}\geq2\sqrt{2}$(Cô-si 4 số)Dấu''='' xra$\Leftrightarrow \begin{cases}a=3b;b(a-b)=1\\ \frac{a-b}{2}=\frac{1}{b(a-b)^{2}}\end{cases}\Leftrightarrow \begin{cases}a=\frac{3\sqrt{2}}{2} \\ b=\frac{\sqrt{2}}{2} \end{cases}$