Proof: We first prove that for any $a, b>0, a+b \leq 1$
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right) \geq\left(\frac{2}{a+b}-1\right)^{2}$$

In fact, this result can be expressed in the following form
$$\begin{array}{l} \frac{1}{a b}-\frac{1}{a}-\frac{1}{b} \geq \frac{4}{(a+b)^{2}}-\frac{4}{a+b} \Leftrightarrow \frac{1}{a b}-\frac{4}{(a+b)^{2}} \geq \frac{1}{a}+\frac{1}{b}-\frac{4}{a+b} \\ \Leftrightarrow \frac{(a-b)^{2}}{a b(a+b)^{2}} \geq \frac{(a-b)^{2}}{a b(a+b)} \Leftrightarrow(a-b)^{2}(1-a-b) \geq 0 \end{array}$$

Returning to the original problem. We can assume $a+b+c+d+e=1$, then we have
$$\prod_{c y c}\left(\frac{a+b}{2}\right) \leq \prod_{c y c}\left(\frac{a+b+c}{3}\right) \Leftrightarrow \prod_{c y c}\left(\frac{a+b+c}{d+e}\right) \geq \frac{3^{5}}{2^{5}} \Leftrightarrow \prod_{c y c}\left(\frac{1}{a+b}-1\right) \geq \frac{3^{5}}{2^{5}}$$

According to $(*)$, we have
$$\left(\frac{1}{d+e}-1\right)\left(\frac{1}{a+b}-1\right) \geq\left(\frac{2}{d+e+a+b}-1\right)^{2}=\left(\frac{2}{1-c}-1\right)^{2}$$

This result indicates
$$\prod_{c y c}\left(\frac{1}{a+b}-1\right) \geq \prod_{c y c}\left(\frac{2}{1-c}-1\right)=\prod_{c y c}\left(\frac{1+c}{1-c}\right)$$

The function $f(x)=\ln (1+x)-\ln (1-x)$ is a convex function, because the second derivative $f^{\prime \prime}(x)=\frac{-1}{(1+x)^{2}}+\frac{1}{(1-x)^{2}} \geq 0$. Therefore, by Jensen's inequality, we have
$$\sum_{c y c} f(a) \geq 5 f\left(\frac{1}{5}\right)=5 \ln \left(\frac{3}{2}\right) \Rightarrow \prod_{c y c}\left(\frac{1+c}{1-c}\right) \geq \frac{3^{5}}{2^{5}} \text { . }$$

Equality holds when $a=b=c=d=e$