Take first the special case where $XYZT$ is a parallelogram with sides parallel to diagonals $AC$, $DB$.
Then $XY+ZT=2EF$. And since $\triangle ZCY$ is right and $F$ bisects $ZY$, then $ZY=2FC$. Likewise $TX=2AE$.
Therefore perimeter $P$ of $XYZT=2AC$.
Now let $X'$ be any other point on $AB$. Join $X'T$ and $X'Y$, and through $X$ draw an ellipse with $T$, $Y$ as foci.
Since $\triangle TAX\sim\triangle YBX$, then$$\angle TXA=\angle YXB$$$AB$ is tangent to the ellipse at $X$ (see Apollonius, Conics III, 48), and all other points $X'$ on $AB$ lie outside the ellipse.
And since by the well-known property of an ellipse$$XT+XY=JT+JY=GH$$but$$JX'+X'Y>JY$$therefore$$X'T+X'Y>XT+XY$$Similarly, taking any other point Z' on $CD$, we show that$$Z'T+Z'Y>ZT+ZY$$Therefore, except in the special case first considered, in a quadrilateral inscribed in a rectangle$$P>2AC$$