*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.

*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

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

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 xyz≠0, x+y+z≠0, 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+y≠0, x+z≠0, 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