Let $U$ be an open, bounded, symmetric and convex subset of $\mathbb{R}^2$ such that $0 \in U$. Define $f: \mathbb{R}^2 \to [0, +\infty)$ as follows: $f(x,y) = \inf \{ \lambda \in \mathbb{R}_+ : (x,y) \in \lambda U \}$.
Prove that $f$ is a norm on $\mathbb{R}^2$.
It is easy to prove that $f(x,y) = 0 \Rightarrow (x,y) = (0,0)$ and that $f(\alpha x, \alpha y) = |\alpha|f(x,y)$. But how can I prove the triangle inequality?
Let $x,y\in \mathbb R^2$ such that $f(x)<a$ and $f(y)<b$. Then, there are $s\in (0,a)$ and $t\in (0,b)$ such that $x\in sU$ and $y\in tU$. Then, $x+y\in sU+tU=(s+t)\left ( \frac{s}{s+t}U+\frac{t}{s+t}U \right )\subseteq (s+t)U,\ $the last inclusion because $U$ is convex. This shows that $f(x+y)\le s+t.\ $To finish, inf over $s$ and $t$.