Post again because this post has bad formatting/not written in $\text{\LaTeX}$. (I'm currently trying to create a collection of Thailand October Camp)

The left inequality is proved in the above link. For the right inequality, the condition is unnecessary. Let $x=\tfrac{ta}{b}$, $y=\tfrac{tb}{c}$, $z=\tfrac{tc}{b}$, then it is equivalent to \[ \sum_{cyc} \frac{bc}{at(bt+c)} \ge \frac{3}{t(t+1)}. \]Indeed, using Cauchy-Schwarz we have \[ \sum_{cyc} \frac{bc}{at(bt+c)} \ge \frac{(bc+ca+ab)^2}{\sum_{cyc} bc \cdot at(bt+c)} = \frac{(bc+ca+ab)^2}{abc(a+b+c)} \cdot \frac{1}{t(t+1)} \ge \frac{3}{t(t+1)}. \]

Quidditch wrote: Let $x, y, z$ be positive real numbers satisfying $x + y + z =\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}$. Prove that $$\frac{3}{2}\leq\frac{3}{\sqrt[3]{xyz}(\sqrt[3]{xyz}+1)}\leq\frac{1}{x(y+1)}+\frac{1}{y(z+1)}+\frac{1}{z(x+1)}.$$ $xy+xz+yz\leq3$ is also true.