For any 4 distinct points $P_{1}, P_{2}, P_{3}, P_{4}$ in the plane, find the minimum value of the ratio
$$
\frac{\sum_{1 \leqslant i<j \leqslant 4} P_{4} P_{j}}{\min _{1 \leqslant i<4} P_{4} P_{j}}
$$

(Wang Jianwei)
Four, first prove the following lemma.
Lemma In $\triangle A B C$, if $A B \geqslant m, A C \geqslant m, \angle B A C = \alpha$. Then $B C \geqslant 2 m \sin \frac{\alpha}{2}$.
Proof of the lemma: Draw the angle bisector $A D$ of $\angle A$. By the Law of Sines, we have
$$
\begin{aligned}
B C= & B D+D C \\
= & A B \frac{\sin \angle B A D}{\sin \angle A D B} \\
& +A C \frac{\sin \angle C A D}{\sin \angle A D C} \\
= & \frac{\sin \frac{\alpha}{2}}{\sin \angle A D B}(A B+A C) \geqslant 2 m \sin \frac{\alpha}{2} .
\end{aligned}
$$

Returning to the solution of the original problem.
Let $m=\min _{1 \leqslant 160^{\circ}$.

So.
$$
\frac{\alpha+\beta}{4} \geqslant 45^{\circ} .0^{\circ} \leqslant \frac{\alpha}{4}<15^{\circ} .
$$

By the lemma, we have
$$
P_{2} P_{4} \geqslant 2 m \sin \frac{\alpha}{2}, P_{1} P_{3} \geqslant 2 m \sin \frac{\beta}{2} .
$$

Then $P_{1} P_{3}+P_{2} P_{4}$
$$
\begin{array}{l}
\geqslant 2 m\left(\sin \frac{\alpha}{2}+\sin \frac{\beta}{2}\right)=4 m \sin \frac{\alpha+\beta}{4} \cdot \cos \frac{\alpha-\beta}{4} \\
\geqslant 4 m \sin 45^{\circ} \cdot \cos 15^{\circ}=2 m\left(\sin 60^{\circ}+\sin 30^{\circ}\right) \\
=(\sqrt{3}+1) m .
\end{array}
$$

Therefore, the inequality holds.
In summary, in all cases, inequality (2) holds.
On the other hand, when $P_{1}, P_{2}, P_{3}, P_{4}$ are the four vertices of a quadrilateral with one interior angle of $60^{\circ}$, the equality in (2) holds.
Therefore, the minimum value of $\frac{k}{m}$ is $5+\sqrt{3}$.