Let $E$ be a vector space over the real (the complex case is interesting too). We consider functions $f:E\rightarrow\mathbb R$ which satisfy the homogeneity property $$f(\lambda x)=|\lambda|\,f(x).$$ In particular, we have $f(0)=0$.
I am interested in the generalized Hlawka inequality (GH$n$) at order $n$. Given $n$ vectors $x_1,\ldots,x_n$, and $I\subset[\![1,n]\!]$, define $x_I=\sum_{i\in I}x_i$. Then (GH$n$) says that $$\sum_{I\subset[\![1,n]\!]}(-1)^{{\rm card}\, I}f(x_I)\le0,\qquad\forall x_1,\ldots,x_n\in E.$$ For instance, (GH1) and (GH2) are respectively $0\le f(x)$ and $f(x+y)\le f(x)+f(y)$; thus, a function $f$ satisfying (GH1) and (GH2) is a norm over $E$. (GH3) is Hlawka inequality, which is satisfied by Euclidian norms and some others, though not by all norms.
Let me point out that (GH$n$) is sharp, in the sense that the equality holds true when either one vector is $0$, or all the vectors are positively collinear. If $n$ is odd, it is also an equality when $\sum_1^nx_i=0$, while if $n$ is even, this choice in (GH$n$) yields (GH$(n-1)$). One may also derive (GH2) from (GH3) by choosing $x_3=x_1$.
First question :
Do Euclidian norms satisfy (GH$n$) for all $n$ ?
Second question :
Conversely, if a norm satisfies (GH$n$) for all $n$, is it Euclidian ?
Third question :
Can we derive (GH$(n-1)$) from (GH$n$) when $n$ is odd ? (True if $n=3$).
Edit. Guillaume, I liked your suggestion. It does give a (new ?) proof of the classical Hlawka inequality ($n=3$). However it fails as soon as $n=4$. A counter-example is $(1,1,1,-2)$, where the sum is $2$. Therefore the answer to the very first question above is No, even in one space-dimension.
