If $X$ is a vector space over field $\mathbb{R}$, then a real-valued, non negative function $ \Vert \cdot , \cdot \Vert$: $X \times X \rightarrow \mathbb{R}$ on $X^2$ is said to be $2$-normed on $X$ if:
i. $\Vert x,y \Vert = 0$ if and only if $x,y$ linearly dependant.
ii. $\Vert x,y \Vert = \Vert y,x \Vert$ for all $x,y \in X$.
iii. $\Vert \alpha x,y \Vert = |\alpha| \Vert x,y\Vert$ for all $\alpha \in \mathbb{R}, x,y \in X$
iv. $\Vert x, y+z \Vert \le \Vert x,y \Vert + \Vert x,z\Vert$ for all $x,y,z \in X$. Geometrically, $\Vert x,y \Vert$ is the area of triangle (or parallelogram) having vertices $0, x, y$.
My problem is this:
Is there any constant $C \in \mathbb{R}$, such that
$$ \Big \Vert \frac{x}{\Vert x \Vert} - \frac{y}{\Vert y\Vert}, \frac{z}{\Vert z \Vert} - \frac{y}{\Vert y \Vert }\Big \Vert \le C \Vert x-y, z-y \Vert$$
If $x,y,z$ are vectors on a unit ball, we can choose $C = 1$. So that, we can obtain the equality. But, I cannot solve this inequality analytically. Thanks in advance.