An Inequality for sides and diagonal of convex quadrilateral from AMM Let $\square ABCD$ be a convex quadrilateral. If the diagonals $AC$ and $BD$ have mid-points $E$ and $F$ respectively, show that:
$$\overline{AB} + \overline{BC} +\overline{CD} + \overline{DA} \ge \overline{AC}+\overline{BD}+2\overline{EF}$$
where, $\overline{XY}$ denotes the length of the line segment $XY$.
The problem was 11841 from May 2015 issue of AMM magazine. While writing complex numbers/vectors for vertices reduces the problem to the well known Hlawka's Inequality for Inner-product spaces, I am interested in purely geometrical solutions.
(It's way past the last date of submission, so I believe it's safe to ask for alternative solutions here.)
Edit: We have the following reformulation, (not that it makes the job any easier though!)

If $H,I,J$ and $K$ are the midpoints of $AD,DC,CB$ and $BA$ respectively, then $\square HIJK$ is a parallelogram and it's easy to see that the diagonals $\overline{HJ}$ and $\overline{IK}$ intersect at $G$, which is the midpoint of $\overline{EF}$ as well.
Then we have the equivalent reformulation of the question:
In a parallelogram $\square HIJK$, whith diagonals intersecting at $G$ and $F$ be an interior point, we need to show:
$$\overline{FH}+\overline{FI}+\overline{FJ}+\overline{FK} > \overline{IJ}+\overline{KJ}+2\overline{FG}$$
 A: The simplest special case is when $\overline{ABCD}$ is a square with side length $a$. In this case, the diagonals are of equal length and given by $a\sqrt{2}$.
$4a \ge 2a\sqrt{2},\quad 4 \ge 2\sqrt{2},\quad$the conjecture holds.
The next simplest case is when $\overline{ABCD}$ is a rectangle of side lengths $a$ and $b$. In this case the diagonals are still equal and given by $\sqrt{a^2 +b^2}$, so 
$2(a+b)\ge 2\sqrt{a^2 + b^2}, \quad a+b\ge \sqrt{a^2 + b^2}, \quad a^2 + 2ab + b^2 \ge a^2 + b^2, \quad 2ab \ge 0, \quad$ and the conjecture holds.
The next case is when $\overline{ABCD}$ is a parallelogram, with sides $a, b$ and interior angles $h, k$. The diagonals can be given by the law of cosines, $d_1 = \sqrt{a^2 + b^2 + 2abcos(h)}$, and $d_2 = \sqrt{a^2 + b^2 + 2abcos(k)}$. These can both be written in terms of one interior angle, since ${h, k} \in [0, \pi]$ and $h=\pi-k$, and $cos(\pi-h) = -cos(h)$. Ie: $d_{1,2} = \sqrt{a^2 + b^2 \pm 2abcos(h)}$. 
$2(a+b) \ge \sqrt{a^2 + b^2 + 2abcos(h)} + \sqrt{a^2 +b^2 -2abcos(h)}$ 
$4(a^2 + 2ab + b^2) \ge a^2 + b^2 + 2abcos(h) + 2\sqrt{(a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))} + a^2 +b^2 -2abcos(h)$
$4a^2 + 8ab + 4b^2 \ge 2a^2 + 2b^2 + 2\sqrt{(a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))}$
$a^2 + 4ab + b^2  \ge \sqrt{(a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))}$
$(a^2 + 4ab + b^2)^2 \ge (a^2 + b^2 + 2abcos(h)(a^2+b^2 -2abcos(h))$
(in the interest of space, I’m gonna skip the foiling and ask you to trust me that this simplifies down to the following:)
$2a^3b+4a^2b^2+2ab^3\ge-a^2b^2cos(h)$
That $cos(h)$ term is bounded in $[-1, 1]$ so worst case scenario, the right hand side of that equality is negative and the inequality is obviously true. Best case scenario, $2a^3b+2a^2b^2+2ab^3\ge 0$ and the conjecture holds.
In all of the above cases, $\overline{EF} = 0$, because for all parallelograms, the diagonals intersect at their midpoints.
The next case is a right trapezoid, with sides $b_1, b_2, h,$ and the last side can be given by $\sqrt{h^2 + (b_2-b_1)^2}$. Assume $b_1 \le b_2$. This is the first case in which the line $\overline{EF}$ will come into play.  In a right trapezoid, the line $\overline{EF}$ lies on the perpendicular bisector of $h$, and its length is given by $\frac{b_2-b_1}{2}$. And the lengths of the diagonals $d_1, d_2$ can be given by $d_1 = \sqrt{h^2 +b_1^2},$ and $d_2=\sqrt{h^2+b_2^2}$. Finally we can write out our inequality as:
$h + b_1 + b_2 + \sqrt{h^2 + (b_2-b_1)^2} \ge \sqrt{h^2 + b_1^2} + \sqrt{h^2 + b_2^2} +2\frac{(b_2-b_1)}{2}$
But alas, in attempting to solve this inequality I realized I would have to square a four term polynomial. I have already shown geometrically that the conjecture holds for all parallelograms. The next cases to consider are this case of the right trapezoid, then the general trapezoid, and finally the general quadrilateral. If you still want a purely geometric solution, perhaps I’ll come back and give it another shot.
