Given a rectangle $ABCD$ in which there is an inscribed quadrilateral $XYZT$, with exactly one vertex on each side of the rectangle, how could I prove that the perimeter of the inscribed quadrilateral is larger then $2|AC|$ (two diagonals)?
I tried to use the triangle inequality, but I can't find the right way to do it.
Let’s suppose that $X$, $Y$, $Z$, $T$ are in $AB$, $BC$, $CD$, $DA$, respectively. Construct the reflection $X_1$ of $X$ through $AD$, the reflection $X_2$ of $X$ through $BC$, and the reflection $X_3$ of $X_1$ through $CD$. Your diagram should look like this.
Now, we have $$XY+YZ+ZT+TX$$ $$=X_2Y+YZ+ZT+TX_1$$ $$\ge X_2Z+ZX_1$$ $$=X_2Z+ZX_3$$ $$\ge X_3X_2.$$ However, since $X_1X_2=2AB$, $X_1X_3=2AD$, $\angle X_3X_1X_2=\angle DAB=90^\circ$, $$\mathop{\bigtriangleup}\!X_1X_2X_3\mathrel{\sim}\mathop{\bigtriangleup}\!ABD,$$ so that $$X_3X_2=2AC,$$ and $XY+YZ+ZT+TX\ge 2AC$, as we wanted. $\blacksquare$
Just mirror your rectangle several times.
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$$P>2AC$$
