### Proof

We place the problem in complex plane, with $A=a,\;$ $B=b,\;$ $C=c,\;$ $D=d,\;$ such that $E=\displaystyle\frac{a+c}{2}\;$ and $F=\displaystyle\frac{b+d}{2}.\;$ We also have $AB=|b-a|,\;$ $BC=|c-b|,\;$ $CD=|d-c|,\;$ $DA=|a-d|,\;$ $AC=|c-a|,\;$ $BD=|d-b|,\;$ $2EF=|d-c+b-a|.\;$ Let's introduce $x=b-a,\;$ $y=c-b,\;$ and $z=d-c,\;$ making $d-a=x+y+z,\;$ $c-a=x+y,\;$ $d-b=y+z,\;$ and $d-c+b-a=z+x.\;$ The problem rewrites as

$|x|+|y|+|z|+|x+y+z|\gt |x+y|+|y+z|+|z+x|.$

But, according to Hlawka's inequality,

$|x|+|y|+|z|+|x+y+z|\ge |x+y|+|y+z|+|z+x|.$

Thus it remains to be shown that the inequality is always strict.

Since $x,y,z,x+y+z\ne 0,\;$ the condition for Hlawka's inequality to become equality can be expressed as

$\displaystyle\frac{x(x+y+z)}{yz}\gt 0,\;\frac{y(x+y+z)}{zx}\gt 0,\;\frac{z(x+y+z)}{xy}\gt 0.$

This would imply, e.g.,

$\displaystyle\frac{\displaystyle\frac{x(x+y+z)}{yz}}{\displaystyle\frac{y(x+y+z)}{zx}}=\left(\frac{x}{y}\right)^2\gt 0,$

with the conclusion that $\displaystyle\frac{x}{y}=\frac{b-a}{c-b}\;$ is a real number, i.e., that $A,B,C$ are collinear, which is impossible.

### Note

The problem discussed above has been taken up by E. A. Weinstein and J. D. Klemm who observed (and proved) that the inequality holds in quadrilaterals not necessarily convex and is always strict as long as the quadrilateral is nondegenerate.

### Acknowledgment

This is the problem 11841 from The American Mathematical Monthly (May 2015) and is due to Leo Giugiuc. Leo has kindly communicated the problem to me, along with a solution.