Grégoire Nicollier
September 27, 2016

The following result improves problem 11841 from The American Mathematical Monthly (May 2015) and http://www.cut-the-knot.org/m/Geometry/HlawkaInConvexQuadrilateral.shtml.

The perimeter of a planar quadrilateral is never shorter than the sum of its diagonals augmented by twice the distance between the midpoints of the diagonals. Equality is attained exactly when two consecutive vertices coincide or three of the oriented sides have the same direction.

Proof. Hlawka's inequality states that

 |x|+|y|+|z|-|x+y|-|y+z|-|z+x|+|x+y+z|\ge0\; \text{for any complex}\;x,\,y,\,z


with equality exactly when one of the four numbers x, y, z, and -(x+y+z) is zero or three of them have the same argument (proof below).

Take now simply

where a, b, c, and d are the vertices of the quadrilateral in order.

Proof of Hlawka's inequality. Let LHS be the left-hand side of Hlawka's inequality. A straightforward calculation shows that

\text{LHS}\cdot\left(|x|+|y|+|z|+|x+y+z|\right)=\sum_{\small\text{cycl}}\left(|x|+|y|-|x+y|\right)\left(|x+y+z|+|-z|-|x+y|\right).

Every term of the cyclic sum is a product of two nonnegative factors by triangle inequalities and the factor multiplying LHS is strictly positive except for x=y=z=0 (but then \text{LHS}=0). It remains to determine when Hlawka's inequality is an equality.

If one of the four numbers x, y, z, and -(x+y+z) is zero or three of them have the same argument, each term of the above cyclic sum vanishes and \text{LHS}=0 follows (or is true in the case x=y=z=0).

To establish that there is no other case of equality, we suppose for the rest of the proof that \text{LHS}=0 with xyz0, x+y+z0, and, say, \arg x different from \arg y and \arg z. We have to show that y, z, and -(x+y+z) have the same argument. In the cyclic sum, the first and third terms have a nonzero first factor, hence the second factor vanishes: x+y+z, -z, and -y have thus the same argument, which implies x+y0, x+z0, z=\lambda(x+y), and y=\mu(x+z) for some real \lambda,\,\mu\in(-1,\,0). As z=-x+y/\mu=\lambda x+\lambda y with \lambda-1, the vectors x and y must be parallel, that is, \arg y=-\arg x. Hence z=\lambda(x+y) is also parallel to x, that is, \arg z=-\arg x. As \lambda\in(-1,\,0), one has \arg(x+y)=-\arg z=\arg x and thus |x|>\max\left(|y|,\,|z|\right). In the first term of the cyclic sum, the second factor

|x+y+z|+|-z|-|x+y|=\left||x|-|y|-|z|\right|+|z|-|x|+|y|
vanishes, that is, |x|\ge|y|+|z|. This implies \arg(x+y+z)=\arg x. Thus y, z, and -(x+y+z) have the same argument and we are done.