How to find the minimum of the expression? Let  $a$, $b$, $c$ be three real positive numbersand $a^2 + b^2 + c^2 =3$. Find the minimum of the expression
$$P = \dfrac{a^2}{b + 2c} +\dfrac{b^2}{c + 2a}+ \dfrac{c^2}{a + 2b}.$$
I tried 
$$\dfrac{a^2}{b + 2c} +\dfrac{b^2}{c + 2a}+ \dfrac{c^2}{a + 2b} \geqslant \dfrac{(a + b+c)^2}{3(a + b + c)}$$
or $$P \geqslant \dfrac{a + b + c}{3}$$

$P\geq \frac{9}{a^2(b+2c)+b^2(c+2a)+c^2(a+2b)}$ by Cauchy-Schwarz inequality.
$a^2b+b^2c+ac^2\leq\sqrt{3(a^2b^2+a^2c^2+b^2c^2)}$ and $2(a^2c+ab^2+bc^2)\leq 2\sqrt{3(a^2b^2+a^2c^2+b^2c^2)}$, again by Cauchy-Schwarz inequality and $P\geq \frac{\sqrt{3}}{\sqrt{(a^2b^2+a^2c^2+b^2c^2)}}$. Arithmetic-geometric inequality gives $(a^2b^2+a^2c^2+b^2c^2)\leq \frac{(a^2+b^2+c^2)^2}{3}=3$, so finally $P\geq 1$.
