Problem 1488

Considering the equality holds for $a=b=c=d$, we try to reduce dimensions, bringing $a$ closer to $b$ and $c$ closer to $d$.
$$\Leftrightarrow \frac{(a+b+c)(a+c+d)(a+b+d)}{(a+b)(a+d)} \cdot \frac{b+c+d}{(b+c)(c+d)} \geqslant\left(\frac{3}{2}\right)^{4}$$

Fix $b, c, d$, and let
$$\begin{array}{l} f(a)=\frac{(a+b+c)(a+c+d)(a+b+d)}{(a+b)(a+d)} \\ f^{\prime}(a)=\frac{\left[\sum_{c y c}(a+b+c)(a+b+d)(c+d)\right](a+b)(a+d)}{[(a+b)(a+d)]^{2}}- \\ \frac{[b(a+d)+d(a+b)](a+b+c)(a+c+d)(a+b+d)}{[(a+b)(a+d)]^{2}}= \\ \frac{g(a)}{[(a+b)(a+d)]^{2}} \\ g(a)=(a+b)(a+c+d)(a+b+d)[(a+d)(b+c)-d(a+b+c)]+ \\ (a+d)(a+b+c)(a+c+d)[(a+b)(a+d)-b(a+b+d)]+ \\ (a+b)(a+d)(c+d)(a+b+c)(a+b+d)>0 \end{array}$$

Thus, $f^{\prime}(a)>0$, so $f(a)_{\text {min }}=f(b)$.
Similarly, we get $f(c)_{\text {min }}=f(d)$.
Therefore, we only need to prove
$$\begin{array}{l} \frac{2 b+d}{2 b} \cdot \frac{b+2 d}{b+d} \cdot \frac{b+2 d}{2 d} \cdot \frac{2 b+d}{b+d} \geqslant\left(\frac{3}{2}\right)^{4} \Leftrightarrow \\ \frac{(2 b+d)(b+2 d)}{\sqrt{b d}(b+d)} \geqslant \frac{9}{2} \Leftrightarrow \\ 4 b^{2}+4 d^{2}+10 b d \geqslant 9 b^{\frac{3}{2}} d^{\frac{1}{2}}+9 d^{\frac{3}{2}} b^{\frac{1}{2}} \end{array}$$

Also, since
$$\begin{array}{l} \left(b^{\frac{1}{2}}-d^{\frac{1}{2}}\right)^{4} \geqslant 0 \Leftrightarrow b^{2}+d^{2}+6 b d \geqslant 4\left(b^{\frac{3}{2}} d^{\frac{1}{2}}+d^{\frac{3}{2}} b^{\frac{1}{2}}\right) \Rightarrow \\ 4 b^{2}+4 d^{2}+10 b d \geqslant 16\left(b^{\frac{3}{2}} d^{\frac{1}{2}}+d^{\frac{3}{2}} b^{\frac{1}{2}}\right)-14 b d \geqslant \\ 9 b^{\frac{3}{2}} d^{\frac{1}{2}}+9 d^{\frac{3}{2}} b^{\frac{1}{2}} \end{array}$$

Thus, the proposition is proved!