$a^2b+a^2c=\frac{a^2b+a^2c}{abc}=\frac{ab+ac}{bc}$$\Rightarrow \frac{bc}{a^2b+a^2c}=\frac{b^2c^2}{ab+ac}$
$\Rightarrow P = \sum\frac{b^2c^2}{ab+ac}$
$\Leftrightarrow P+\frac{ab+bc+ca}2=(\frac{b^2c^2}{ab+ac}+\frac{ab+ac}{4})+(\frac{c^2a^2}{bc+ba}+\frac{bc+ba}{4})+(\frac{a^2b^2}{ca+cb}+\frac{ca+cb}{4}) \overset{cosi}{\ge} bc+ca+ab$
$\Leftrightarrow P \ge \frac{ab+bc+ca}{2} \ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac 32$