How prove that $ AB +BC + CD + DA\leq 2 + 4\sin\frac {\pi}{12}$ for convex quarilateral $ ABCD$? Let $ ABCD$ be convex quarilateral and $ \max\{AB,BC,CD,DA,AC,BD\}\leq 1$. How prove that
$ AB +BC + CD + DA\leq 2 + 4\sin\frac {\pi}{12}$?
 A: Assume that $ABCD$ has the maximum possible perimeter.
We can also assume that both the diagonals $AC$ and $BD$ have unit length, since otherwise we can stretch a little one of the endpoints, increasing the perimeter without violating the constraints.
We can assume that $A$ lies in the region $\Omega_A$ of points $P$, above $BD$, for which $PB\leq 1,PD\leq 1$. 
Call $\Omega_C$ the symmetric of $\Omega_A$ with respect to $BD$ and $U$,$V$ the vertices of the region $\Omega=BD\cup\Omega_A\cup\Omega_C$. Additionally we can assume  $AB\leq AD$ and $AU\geq 1$ (otherwise, exchange the roles of $A,U$ and $C,V$).

Since the sum of the distances from two points is a convex function, and convex functions attain their maximum on the boundary of closed sets, we can further assume that $C$ lies on the boundary of $\Omega$, such that $A$ and $C$ are on the same half-plane with respect to the $UV$ line.
If $A$ differs from the $W$ point in the picture and does not lie on $\partial\Omega$, there is a point $A'\in\Omega_A$ such that the perimeter of $A'BCD$ is strictly greater than the perimeter of $ABCD$. Consider the ellipse $\Gamma$ through $A$ with foci in $B$,$D$: since this ellipse is the locus of points $Q$ for which $QB+QD=AB+AD$, we can take a point $A'\in\Omega_A$, outside $\Gamma$, on the circle through $A$ having center $C$. By this way, $A'C=AC$ and $A'B+A'D>AB+AD$.
Since this must not happen, either $C\equiv U$ and $ABCD$ is the polygon $BUDW$ depicted above (having perimeter $2+4\sin\frac{\pi}{12}$), or both $A$ and $C$ lie on the boundary of $\Omega$.
This is a way easier optimization problem: twice the area of $ABCD$ is just the sine of the angle between $AC$ and $BD$, so the maximum of the area occurs when $AC\perp BD$, and it is not difficult to see that the solution is given by a polygon that is similar to the previous $BUDW$.
