Let $\|\cdot\|$ be a norm on a vector space $V$, and define, for each ordered pair of vectors, the scalar $d(x,y) = \|x-y\|$, called distance between $x$ and $y$. Prove the following results for all $x,y,z\in V$.
$d(x,y) \leq d(x,z)+d(z,y)$
I tried to prove this in different ways, I tried to prove $\|x-y\|^{2}\leq (||x-z|| + ||z-y||)^{2}$ and expanded everything out, move this and that around, but none of my tactics work. Please give me some insight or trick of how to prove this. Thank you.