To prove: Given the conditions, there exist four positive real numbers $x, y, z, t$ satisfying
$$a=\frac{r y}{x}, b=\frac{r z}{y}, c=\frac{r t}{z}, d=\frac{r x}{t}$$

The inequality then becomes the following form

We need to prove $A+\left(r^{2}-1\right) B \geq \frac{4\left(r^{2}+1\right)}{r+1}$, where
$$A=\sum_{c y c} \frac{x+z}{r y+x} ; \quad B=\sum_{c y c} \frac{z}{r y+x}$$

By the AM-GM inequality, we have
$$\begin{array}{l} 4 r \sum_{c y c} x y+8(x z+y t)=[4(r-1)(x+z)(y+t)]+4[(x+z)(y+t)+2(x z+y t)] \\ \leq(r-1)\left(\sum_{c y c} x\right)^{2}+2\left(\sum_{c y c} x\right)^{2}=(r+1)\left(\sum_{c y c} x\right)^{2} \end{array}$$

According to the Cauchy-Schwarz inequality, and noting that $r \geq 1$, we have
$$\begin{array}{l} A=(x+z)\left(\frac{1}{r y+x}+\frac{1}{r t+z}\right)+(y+t)\left(\frac{1}{r x+y}+\frac{1}{r z+t}\right) \geq \frac{4(x+z)}{x+z+r y+r t}+\frac{4(y+t)}{y+t+r x+r z} \\ \geq \frac{4(x+y+z+t)^{2}}{(x+z)^{2}+(y+t)^{2}+2 r(x+z)(y+t)} \geq \frac{8}{r+1} \\ B \geq \frac{(x+y+z+t)^{2}}{z(r y+x)+t(r z+y)+x(r t+z)+y(r x+t)} \geq \frac{(x+y+z+t)^{2}}{r(x y+y z+z t+t x)+2(x z+y t)} \geq \frac{4}{r+1} \end{array}$$

Thus, the inequality is proven. The equality holds when $a=b=c=d=r$