I am trying to figure out the conditions under which the inequality
$$|x|>|z|$$
holds, where $z$ is a convex combination of $x,y \in \mathbb{R}$:
$$z= \pi x + (1-\pi) y \quad , \quad \pi \in (0,1).$$
Surely, $ x \ne y $, but which other conditions need to be met?
My first intuition was $y=0$, which is a sufficient condition. A second set of conditions seems to be $|x|>|y|$ and $sgn(x)=sgn(y)$, because then it must hold that
$$|x|>|z|>|y|.$$
Are these the two solutions or is there a more general set? How should I best approach this problem? My approach seems a bit unstructured.
edit: please suggest better tags for this problem too
I will prove:
$|x|>|z|$ for all $\pi\in(0,1)$ if and only if $|x|>|y|$ or $x=-y\neq 0$
(You should be able to see this result is true by drawing a diagram.)
Sufficiency:
If $|x|>|y|$ then for all $\pi\in(0,1)$
$$|x|>\pi|x|+(1-\pi)|y|=| \pi x|+|(1-\pi)y|\geq |\pi x + (1-\pi) y|=|z|$$
where the second inequality is from the triangle inequality.
If $x=-y\neq 0$ then for all $\pi\in(0,1)$:
$$|z|=|\pi x-(1-\pi)x|=|(2\pi-1)x|=|2\pi-1||x|<|x|.$$
Necessity:
If $x=y$ then $|x|=|z|$.
Finally we show that if $|x|<|y|$ then there exists some $\pi\in(0,1)$ such that $|x|<|z|$. Let
$$f(\pi)=|x|-|\pi x+(1-\pi)y|$$
Suppose $|x|<|y|$, then we have
$$f(0)=|x|-|y|<0$$
and so by continuity of $f$ in $\pi$, we have that $f(\pi)=|x|-|z|<0$ for small $\pi>0$.