Here is a solution based on trigonometry (though, in a somewhat less elegant wording, it can be translated step-for-step to a Euclidean argument).

First, let's prove the following lemma.

**Lemma:** For any triangle *ABC* it is true that the triangle
*A'B'C'* satisfying that the angle *ABC* equals the angle
*A'B'C'* and that *|AB|+|BC|=|A'B'|+|B'C'|* and that
*|A'B'|=|B'C'|* satisfies *AC*≥*A'C'*.

Proof of the lemma:

The general edge-length relation in triangles is that

To minimise *|AC|* we would need to minimise the right-hand side of this
equation, and therefore also the left-hand side. Because of the condition
*|AB|+|BC|=|A'B'|+|B'C'|*, we can treat

as a constant. Subtracting from this constant the left-hand side of the first displayed equation we get that we need to maximise

Because angle *ABC* equals angle *A'B'C'*, we can treat it also
as a constant. Furthermore, cos(*ABC*) is bounded from below by -1, so we
have that we need to maximise *|AB||BC|* while not changing their
sum. The solution is therefore to make them equal to each other. This is the
definition of *A'B'C'*.

Q.E.D.

(An alternate and perhaps more elegant proof of the same lemma is to see that
(|*AB*|+|*BC*|)sin(*ABC*/2) is the length of the projection of
*AC* onto the perpendicular of *ABC*'s angle bisector, which is a
lower bound on the length of *AC* itself. *A'C'* equals this bound.)

With this lemma, it is possible to prove the main theorem.

Proof:

First, note that we can assume that *ABCD* is convex. (Suppose, on the
contrary that, without loss of generality, angle *ABC* is greater than
180deg. Then by taking *AB'CD*, where *B'* is the mirror image of
*B* on the other side of *AC*, we can construct a quadrilateral
with the same side lengths but longer diagonal sum. By repeating this at most
twice, we get a convex quadrilateral.)

With this assumption, the two diagonals of *ABCD* intersect at a point
inside the quadrilateral. Let us call this point *E*,
let angle *AEB* be called θ and let θ/2 be ψ.

By the lemma, *|AB|* can be bounded from below by
(*|AE|*+*|EB|*)*sin(ψ).
Similarly, *|CD|* can be bounded from below by
(*|CE|*+*|ED|*)*sin(ψ).
Altogether, *|AB|+|CD|* can be bounded from below by
(*|AC|*+*|BD|*)*sin(ψ).

For exactly the same reason, *|BC|*+*|DA|* can be bounded from below
by (*|AC|*+*|BD|*)*sin((180-θ)/2) =
(*|AC|*+*|BD|*)*cos(ψ).

The largest of *|AB|* and *|CD|* can be bounded from below by
(*|AC|*+*|BD|*)*sin(ψ)/2, and
the largest of *|BC|* and *|DA|* can be bounded from below by
(*|AC|*+*|BD|*)*cos(ψ)/2.

So the sum of the largest three of the edges can be bounded from below by
(*|AC|*+*|BD|*)*max(sin(ψ)+cos(ψ)/2, sin(ψ)/2 + cos(ψ)).

To prove the theorem, we need to show that the minimum of max(sin(ψ)+cos(ψ)/2, sin(ψ)/2 + cos(ψ)) is greater than 1.

The minimum of this expression is attained at the extremes: ψ=0 or ψ=90deg. This minimum is 1 (although, because ψ can never actually attain either extreme value, neither can an exact ratio of 1 be attained by any quadrilateral).

Q.E.D.

Note that a ratio arbitrarily close to 1 can be attained when *ABCD* is
a very long and narrow rectangle.

Last note:

For all the geometry purists in the audience, I decided to also publish this
month one solution that is purely classic Euclidean geometry. For this, I am
using the solution sent to me by Daniel Bitin. It appears
here exactly as it was sent to me. I haven't
altered a word. Thanks, Daniel!