Minimum sum in quadrilateral Let $ABCD$ be a convex quadrilateral. Also, let $E$ be the midpoint of $AC$ and $F$ the midpoint of $BD$ and $M$ the midpoint of $EF$. The circle $\Gamma$ with center $M$ has radius $r$, such that all the points $A,B,C,D$ lie outside the circle. Prove that, for any point $P$ in the interior of $\Gamma$ $$PA+PB+PC+PD>4r$$
I tried using complex number. One can easily see that $m=\frac{a+b+c+d}{4}$ and $PA+PB+PC+PD=|p-a|+|p-b|+|p-c|+|p-d|$. 
I also suspect that the minimum is achieved when $P$ is at the intersection $O$ of the diagonals, or at the intersection of $MO$ and $\Gamma$ when $O$ is outside the circle, but didn't succeed in proving it.
 A: For starters, let us prove your last claim.

If $ABCD$ is a convex quadrilateral and $PA+PB+PC+PD$ is minimal, $P=AC\cap BD$.

If we assume that $Q$ makes $QA+QB+QC+QD$ minimal we may consider the ellipse $\Gamma_{AB}$ with foci at $A,B$ through $Q$ and the ellipse $\Gamma_{CD}$ with foci at $C,D$ through $Q$. With our assumptions $\Gamma_{AB}$ and $\Gamma_{CD}$ have to be tangent, otherwise it would be possible to simultaneously decrease both $QA+QB$ and $QC+QD$ by moving $Q$ along the segment joining the two intersection points. By the optical property of the ellipse, the commont tangent at $Q$ must be perpendicular to both the angle bisector of $\widehat{CQD}$ and the angle bisector of $\widehat{AQB}$: it follows that the minimum is unique and it is achieved at $AC\cap BD$.

Your question now is much simplified. If $P=AC\cap BD$ lies inside $\Gamma$, it is enough to show that $BD=PB+PD\geq 2r$ and $AC=PA+PC\geq 2r$, with at least one inequality being tight.
A: Lemma: If $ABCD$ is a convex quadrilateral and $PA+PB+PC+PD$ is minimal, then $P=AC \cap BD$.
Proof: Let $Q=AC \cap BD$.
By the triangle inequality in triangle PAC we have
$$PA+PC \geq AC=QA+QC$$
with equality if and only if $P \in (AC)$.
Same wway
$$PB+PD \geq BD=QB+QD$$
with equality if and only if $P \in (BD)$.
This shows that 
$$PA+PB+PC+PD \geq QA+QB+QC+QD$$
with equality if and only if $P\in (AC)\cap (BD)=\{Q\}$.
A: Let (1) A’ be the midpoint of AB. B’ C’ and D’ are similarly defined; and (2) A’’ be the midpoint of A’B’. B’’ C’’ and D’’ are similarly defined.
Fact-1) B’’MD’’ and A''MC'' are a straight lines.

By midpoint theorem, BB’EC’ is a parallelogram with BB’’ = B’’E. This, together with the fact that FM = ME, makes MB’’ // BF. Similarly MD’’ // DF. Therefore, B’’MD’’ is a straight line. The same is true for A’’MC’’.

Fact-2) PA + PB + PC + PD > A’B’ + B’C’ + C’D’ + ‘DA’

By the triangular inequality and midpoint theorem, we have:-
PA + PB + PC + PD = (PA + PC) + [PB + PD] > (AC) + [BD] = (B’C’ + A’D’) + [A’B’ + C’D’].

Construction: Thro’ M, draw D’’MB’’//BD and draw A’’MC’’//AC. Then, form the parallelogram A’’B’’C’’D’’. 
Fact-3) (A’B’) + B’C’ + C’D’ + D’A’ > A’’ B’’ + B’’C ‘’+ C’’D’’ + D’’A’’
This is obviously true just by visual inspection.

Fact-4) A’’ B’’ + B’’C ‘’+ C’’D’’ + D’’A’’ = MA’ + MB’ + MC’ + MD’.

By the midpoint theorem, this is true by considering the diagonals of A’B’C’D’.

Result follows because each of the MA’ (and MB’…. ) is longer than the radius of the red circle, provided that the red circle is bounded by the sides of ABCD.
