High school olympiad homogeneity $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$

Question

Problem: Prove for $a,b,c \in \mathbb{R+}$\begin{split} \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1 \end{split}

Solution: Assume $x = b+2c, y = c+2a, z = a+2b$. Rewrite left side to \begin{split} \frac{4y+z-2x}{9}+\frac{4z+x-2y}{9}+\frac{4x+y-2z}{9} = \end{split} \begin{split} = \frac{4y+z-2x}{9}+\frac{4z+x-2y}{9}+\frac{4x+y-2z}{9} = \end{split} \begin{split} = \frac{3(x+y+z)}{9} = \frac{x+y+z}{3} = \end{split} \begin{split} = a+b+c\end{split}

Since LS of inequality is homogenous we can assume $abc=1$. From AM-GM inequality we have \begin{split} a+b+c\geq 3\sqrt[3]{abc} \geq 3 \geq 1 \end{split}

Quesiton: the proof is obviously wrong, after long research I have no idea how homogeneity works, any feedback would be also appreciated.

Answer 1

You made an error in the substitution (check the denominator).

The inequality should be equivalent to

$$ \sum \frac{ 4y+z - 2x} { 9x} \geq 1 . $$

This in turn is equivalent to

$$ \sum \frac{ 4y+z} { x } \geq 15. $$

Can you take it from here?