I'd like to know when the equality holds in the following inequality
$$
| x - y |^a \le | x - z |^a + | y - z |^a.
$$


More precisely, for which points $x = (x\_1, x\_2)$, $y = (y\_1, y\_2)$ and $z = (z\_1, z\_2)$ in the plane does hold
$$
| x - y |^a = | x - z |^a + | y - z |^a
$$
where $a \in (0,1)$?


Notice that when $a=1, z$ can be any point in the line segment joining $x$ to $y$. That is, $z=tx+(1-t)y$ with $t \in (0,1)$. My conjecture is that for $a \in (0,1)$, $z$ must be in curve like cusp instead of a line joining $x$ and $y$.


In other words, is possible to find out a expression like $z=tx+(1−t)y$
as in the case $a=1$?


I tried the particular case: graph $1=(x^2+y^2)^{1/2}+((x-1)^2+y^2)^{1/2}$. This is the case $x=(0,0)$, $y=(1,0)$ and $a=1/2$. I tried in geogebra and wolfram alpha, however I can not see any graph. I do not have softwares like mathematica. Can you help me.


Any help is good. Reference, direct proof, a picture or a link.



When does equality hold in a specific triangle inequality?
For $a\in(0,1)$ and any real $u,v\ge0$,
$$(u+v)^a\le u^a+v^a, \tag{1}\label{1}$$
with the equality if and only if $u=0$ or $v=0$.


Now,
$$|x-y|^a=|(x-z)+(z-y)|^a\le(|x-z|+|z-y|)^a \\ 
\le|x-z|^a+|z-y|^a
=|x-z|^a+|y-z|^a.$$
So, if $|x-y|^a=|x-z|^a+|y-z|^a$, then $(|x-z|+|z-y|)^a 
=|x-z|^a+|z-y|^a$, so that we have the equality in \eqref{1} with $u=|x-z|$ and $v=|z-y|$, which implies $u=0$ or $v=0$, that is, $x=z$ or $y=z$.


Vice versa, if $x=z$ or $y=z$, then clearly $|x-y|^a=|x-z|^a+|y-z|^a$.


We conclude that
$|x-y|^a=|x-z|^a+|y-z|^a$
if only if $x=z$ or $y=z$.


