Simple Triangle-like Inequality for a general norm I came up with this inequality as I was thinking about norms. I don't know if it's true though.
Given three points $x,y,z$ in a normed space $(X,\|\cdot\|)$, is it true that if
$$
\|x-y\|\le\|y\|
$$
and
$$
\|x-z\|\le\|z\|
$$
then
$$
\|x-(y+z)\|\le\|y+z\| ?
$$
In other words, for a given vector $x$ in a normed space $X$, is the set $K:=\{y:\|x-y\|\le\|y\|\}$ additive?
First, if it were true, it almost looks like we could change the $y$ and $z$ around so that, e.g. the 1st and 3rd inequalities imply the 2nd. As it is, given any arrangement $x+y=z$ (none of which is $0$), we can tell which is which by simply adding.
I thought proving $X=\mathbb{R}$ would easily turn into $X=l_p$, but that seems naive.
Inner Product Space
The first two inequalities reduce to
$$
\langle x,x\rangle\le 2\langle y,x\rangle
$$
and
$$
\langle x,x\rangle\le 2\langle z,x\rangle
$$
Adding these together gets
$$
\|x\|^2\le \langle y+z,x\rangle\le 2\langle y+z,x\rangle
$$
Working the same algebra backwards then gets the third inequality.
Other norms?
Note that if $y\in K$, then $cy\in K$ for all $c\ge 1$. So there is a curve/surface in $X$, symmetric about $0$ (AKA odd), that divides $X$ into two parts, one part being $K$. It looks like if the unit ball $B$ of $X$ is oblong and $x$ is not perpendicular to an axis of symmetry of B, the curve/surface will not be linear and hence not convex (symmetric about $0$ and convex implies linear).
So, the factor of $2$ cannot be dismissed for general $X$ as it was for inner product spaces (?). I feel like that argument says the 2 cannot be dismissed for some inner product spaces as well. Mistake?
Thanks!

The inequality can fail for $X=\Bbb R^2$ endowed with $\|\cdot\|_\infty$ norm. Indeed, let $x=(1,0)$, $y=(-1,2)$, and $z=(-1,-2)$. Then $\|x-y\|=\|y\|=\|x-z\|=\|z\|=2$, but $\|x-(y+z)\|=3>2=\|y+z\|$.
