45. 18 $k \star$ Find the smallest real number $\lambda$ such that the inequality
$$
5(a b c+a b d+a c d+b c d) \leqslant \lambda a b c d+12
$$

holds for any positive real numbers $a, b, c, d$ satisfying $a+b+c+d=4$.
When $a=b=c=d=1$, we get $\lambda \geqslant 8$. Below, we prove the inequality
$$
5(a b c+a b d+a c d+b c d) \leqslant 8 a b c d+12
$$

for any positive real numbers $a, b, c, d$ satisfying $a+b+c+d=4$.
Let $f(a, b, c, d) - 5(a b c + a a b d + a c d + b c d) - 8 a b c d$
$$
= a b(5 c + 5 d - 8 c d) + 5 c d(a + b).
$$

If $f(a, b, c, d) > 12$, then $5 c + 5 d > 8 c d$. Otherwise,
$$
f(a, b, c, d) = 5 c d(a + b) \leqslant 5\left(\frac{(a + b) + c + d}{3}\right)^{3} = 5\left(\frac{4}{3}\right)^{3} < 12.
$$

Thus, by the AM-GM inequality, we have
$$
\begin{aligned}
f(a, b, c, d) & = a b(5 c + 5 d - 8 c d) + 5 c d(a + b) \\
& \leqslant \left(\frac{a + b}{2}\right)^{2}(5 c + 5 d - 8 c d) + 5 c d(a + b) \\
& = f\left(\frac{a + b}{2}, \frac{a + b}{2}, c, d\right),
\end{aligned}
$$

Therefore, by symmetry, we get
$$
\begin{aligned}
f(a, b, c, d) & \leqslant f\left(\frac{a + b}{2}, \frac{a + b}{2}, c, d\right) \\
& \leqslant f\left(\frac{a + b}{2}, \frac{a + b}{2}, \frac{c + d}{2}, \frac{c + d}{2}\right) \\
& \leqslant f\left(\frac{a + b}{2}, 1, 1, \frac{c + d}{2}\right) \leqslant f(1, 1, 1, 1) = 12,
\end{aligned}
$$

Thus, inequality (1) is proved.
Therefore, the minimum value of $\lambda$ is 8.