<s>[INST] 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?
 [/INST] Very similar to Nesbitt's inequality. If we set $A=b+2c,B=c+2a,C=a+2b$, we have $ 4A+B-2C = 9c $ and so on, and the original inequality can be written as:
$$ \frac{4B+C-2A}{9A}+\frac{4C+A-2B}{9B}+\frac{4A+B-2C}{9C} \geq 1 $$
or:
$$ \frac{4B+C}{A}+\frac{4C+A}{B}+\frac{4A+B}{C} \geq 15 $$
that follows from combining $\frac{B}{A}+\frac{A}{B}\geq 2$ (consequence of the AM-GM inequality) with $\frac{B}{A}+\frac{C}{B}+\frac{A}{C}\geq 3$ (consequence of the AM-GM inequality again).
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