hmmm... It maybe posted before. But I think we need to expand the following inequality: \[ \sum_{\text{cyclic}}(x+z)(y+z)(z+t)(t+x) \ge 4(x+y)(y+z)(z+t)(t+x) \] sorry, I'm not brave enough.

Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{t}{z}, d = \frac{x}{t}$, then the inequality becomes $\frac{x+z}{x+y} + \frac{y+t}{y+z} + \frac{z+x}{z+t} + \frac{t+y}{t+x} \geq 4$ $\frac{x+z}{x+y} + \frac{z+x}{z+t} =(z+x) (\frac{1}{x+y} + \frac{1}{z+t}) \ge \frac{4(z+x)}{x+y+z+t}$ Similarly, $\frac{y+t}{y+z} + \frac{t+y}{t+x} \ge \frac{4(y+t)}{x+y+z+t}$ Adding the last two together to yield the desired result.

let $a=\frac{y}{x},b=\frac{z}{y},c=\frac{w}{z},d=\frac{x}{w}$ $LHS=\sum_{cyc}\frac{x+z}{x+y}\ge((x+z)+(y+w))\frac{4}{x+y+z+w}=4=RHS$

but it's divinely weird that sometimes two substitutions: $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ and $a=\frac{y}{x},b=\frac{z}{y},c=\frac{x}{z}$ lead to entirely different results!