Prove that $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$ 
For three positive real numbers $a,b,$ and $c$, prove that $$\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b} \geq 1.$$

Attempt
Rewritting we obtain $\dfrac{2 a^3+2 a^2 b-3 a^2 c-3 a b^2-3 a b c+2 a c^2+2 b^3+2 b^2 c-3 b c^2+2 c^3}{(a+2b)(2a+c)(b+2c)} \geq 0$. Then to I  proveed to use rearrangement, AM-GM, etc. on the numerator?

Use Cauchy–Schwarz's inequality as follows :
$$(a(b+2c)+b(c+2a)+c(a+2b)) \left (\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \right ) \geq (a+b+c)^2$$ and then :
$$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}  \geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$
Now it's easy to see that $$\frac{(a+b+c)^2}{3(ab+bc+ca)} \geq 1$$ this being equivalent with $$a^2+b^2+c^2 \geq ab+bc+ca$$ which is in turn equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$$
(after a multiplication with $2$ and rearranging )
