Help to Prove Convex Quadrilateral Problem I'd appreciate help with solving this convex quadrilateral riddle. It appears to hold true in all of my test cases, but I'm not sure how I would go about writing a proof for it.

Let $UVWX$ be a convex quadrilateral. 
  Let $Y$ be the midpoint of $UW$, 
  and let $Z$ be the midpoint of $VX$. 
  Show that $|UV| + |VW| +  |WX| + |XU| \geq |UW| + |VX| + 2|YZ|$ 
  where $|UV|$ denotes the distance from $U$ to $V$.

 A: Since $UVWX$ is convex, all its interior angles are $\le 180^\circ$.
Also
$\text{"Z is midpoint of VY"} \implies 2|YZ| = |VY| \tag{1}$ 
We have the following:


*

*$\angle{UVX} \le \angle{UVW}$ with equality only if $V,X,W$ are collinear 

*$\angle{UVW} \le 180^\circ$ with equality only if $U,V,W$ are collinear.


So $\angle{UVX} < 180^\circ$ unless $U,V,W,X$ are collinear. Such a shape is not a quadrilateral.
Thus we can draw a point $T$ such that $UVXT$ is a parallelogram. Then it follows by similar triangles that $VY$ lies on the diagonal and $|VT| = 2|VY|$. By the triangle inequality applied to $\Delta{VUT}$ we have:
$|VT| \le |VU| + |UT|$ which can be rearranged to $|VT| \le |UV| + |VX|$
Hence
$2|VY| \le |UV| + |VX| \tag{2}$ 
Therefore, it would be sufficient to prove
$|UV| + |VW| +  |WX| + |XU| \ge |UW| + |VX| + \frac{1}{2}(|UV| + |VX|)$
or equivalently
$\frac{1}{2}|UV| + |VW| +  |WX| + |XU| \ge |UW| + \frac{3}{2}|VX| \tag{3}$
But this fails in the case of a unit square where $|UV| = |VW| = |WX| = |XU| = 1, |UW| = |VX| = \sqrt2$ because $3.5 < 2.5 \sqrt2$.
So using this approach, we are at an impasse. Asymmetry considerations alone would make me suspicious of (3) and even of the original inequality to be proved.
