Find the error in my workings for the proof of $\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)\ge 1$ Here are my workings...
By Cauchy-Schwarz,
$$\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)\left( \frac{x+2y}{x} + \frac{y+2z}{y} + \frac{z+2x}{z}\right) ≥ (1+1+1)^2$$
$$\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)(3+2(y/x + z/y + x/z))≥ 9 \tag{1}$$
However, by AM-GM,
$y/x + z/y + x/z ≥ 3,$ meaning that there is no maximum value for $y/x + z/y + x/z.$
Hence, $3+2(y/x + z/y + x/z)$ has no maximum value and we can't find the minimum value of $\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)$ using equation $(1)$?
Please help me spot the error and enlighten me on what's the correct way to solve it (Using AM-GM or Cauchy-Schwarz inequality)

Observe,$$\sum_{\text{cyclic}}\frac{x}{x+2y}=\sum_{\text{cyclic}}\frac{x^2}{x^2+2xy}$$
Using Titu's Lemma,
$$\sum_{\text{cyclic}}\frac{x^2}{x^2+2xy}\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+zx)}=1$$
$$\therefore\;\; \frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\ge 1 $$
