Proving a function $p:V→[0,\infty)$ forms a norm on V if and only if the unit ball $\{{x∈V|ρ(x)<1}\}$ is convex I have a function $p:V→[0,\infty)$ where V is a vector space and we know $p$ to be positive, absolutely homogeneous and non-degenerate (i.e.: we know $p$ satisfies all norm conditions other than the triangle inequality). I need to prove that $p$ forms a norm on $V$ (i.e.: satisfies the triangle inequality) if and only if the unit ball $\{{x∈V|p(x)<1}\}$ is convex. I've shown that if the unit ball is convex, we have $tx+(1−t)y<1$ for $x,y∈V,t∈[0,1]$. Buy am unsure as to where to go from there, or if that's even useful.
I've seen a similar question like this on here, but it is for $V=R^2$,with the proof taking $x∈V$ to be a scalar, and I don't know if I can do that for a general vector space.
Any help is appreciated.

Some hints: If you knew that the closed unit ball $\{ x \in V \mid \rho(x) \le 1 \}$ was convex, then for nonzero vectors $x$ and $y$ you could use the convexity to conclude:
$$ \rho \left( \frac{\rho(x)}{\rho(x) + \rho(y)} \cdot \frac{x}{\rho(x)} + \frac{\rho(y)}{\rho(x) + \rho(y)} \cdot \frac{y}{\rho(y)} \right) \le 1. $$
Now, as for how you show the closed unit ball is convex if the open unit ball is: suppose $\rho(x), \rho(y) \le 1$.  Then for any $\epsilon \in (0, 1)$, we have $\rho(\epsilon x) < 1$ and $\rho(\epsilon y) < 1$, so $\epsilon \rho(tx+(1-t)y) = \rho(t \cdot \epsilon x + (1-t) \cdot \epsilon y) < 1$, and so $\rho(tx + (1-t)y) < \frac{1}{\epsilon}$.
