= Solution
Let $A$ and $C$ be opposite interior angles of a convex quadrilateral with side lengths $a,b,c,d$, and let $s=(a+b+c+d)/2$. The <Bretschneider formula> gives its area $K$ as
$$
K^2=(s-a)(s-b)(s-c)(s-d)
-abcd\cos^2\frac{A+C}{2}.
$$
The fixed side lengths fix the first term, while the second term is nonpositive. Hence
$$
K^2\leq(s-a)(s-b)(s-c)(s-d),
$$
with equality exactly when $A+C=\pi$. Opposite angles are supplementary exactly when the quadrilateral is a <cyclic quadrilateral>. The given $Q$ is cyclic, so it attains this upper bound and
$$
\boxed{\operatorname{Area}(Q)\geq\operatorname{Area}(Q')}.
$$
A crossed or concave competitor can be uncrossed or reflected across a diagonal without changing its side lengths and without decreasing its unsigned area, so the same bound applies. Apart from degenerate coincidences, equality holds precisely when $Q'$ is also cyclic with the stated cyclic ordering of its side lengths.
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