Problem 1448

$A D E F$ are all parallelograms.

Connecting $B F$ and $A E$, it is clear that $B C A F$ is also a parallelogram, thus, $\square$
$$A F=E D=B C, E F=A D, E B=C D, B F=A C$$

In quadrilaterals $A B E F$ and $A E B D$, by Ptolemy's inequality we get
$$\begin{array}{l} A B \cdot E F+A F \cdot B E \geqslant A E \cdot B F \\ \quad B D \cdot A E+A D \cdot B E \geqslant A B \cdot E D \end{array}$$

That is
$$\begin{array}{l} A B \cdot A D+B C \cdot C D \geqslant A E \cdot A C \\ B D \cdot A E+A D \cdot C D \geqslant A B \cdot B C \end{array}$$

Thus, from equations (1) and (2) we can obtain
$$\begin{array}{l} D A \cdot D B \cdot A B+D B \cdot D C \cdot B C+D C \cdot D A \cdot C A= \\ D B(A B \cdot A D+B C \cdot C D)+D C \cdot D A \cdot C A \geqslant \\ D B \cdot A E \cdot A C+D C \cdot D A \cdot A C= \\ A C(B D \cdot A E+A D \cdot C D) \geqslant \\ A C \cdot A B \cdot B C \end{array}$$

Therefore, the inequality is proved, and the equality holds if and only if the equalities in equations (1) and (2) hold simultaneously, i.e., the equality holds if and only if $A B E F$ and $A E B D$ are both cyclic quadrilaterals. This means the equality holds if and only if $A F E B D$ is a cyclic pentagon. Since $A F E D$ is a parallelogram, the condition is equivalent to $A F E D$ being a rectangle (i.e., $A D \perp B C$) and $\angle A B E=\angle A D E=90^{\circ}$, which is also equivalent to $A D \perp B C$ and $C D \perp A B$. Therefore, the necessary and sufficient condition for the equality in the original inequality to hold is that $D$ is the orthocenter of $\triangle A B C$.