Let $X$ be a vector space over $\mathbb K = \mathbb C$ or $\mathbb K =\mathbb R$ and $A \subseteq X$ absorbing.
I want to prove that if $A$ is convex, the Minkowski functional
$$\mu_A:X\to \mathbb R_0^+, \quad \mu_A(x) := \inf\{t \in \mathbb R^+:t^{-1}x\in A\}$$
satisfies the triangle inequality, i.e.
$$\forall x,y\in X: \mu_A(x+y) \leq \mu_A(x)+\mu_A(y).$$
My attempt: I already proved the result, that if a $A$ convex, then $$\forall x\in X \,\forall s>\mu_A(x):s^{-1}x \in A. \quad (1)$$ Now first assume, that $\mu_A(x) = \mu_A(y) =: r$. From $(1)$, we know that for every $\epsilon > 0$ we have $$(r+ \varepsilon)^{-1}x \in A, (r+ \varepsilon)^{-1}y \in A.$$ Since $A$ is convex, $$\frac{1}{2}(r+ \varepsilon)^{-1}x + \frac{1}{2}(r+ \varepsilon)^{-1}y = (r+ \varepsilon)^{-1}(\frac{1}{2}x+\frac{1}{2}y) \in A.$$ By definition that gives us $$\frac{1}{2}\mu_A(x+y)=\mu_A\left(\frac{1}{2}x + \frac{1}{2}y\right) \leq r+\epsilon = \mu_A(x) + \mu_B(x) + \epsilon,$$ proving this special case. Now I tried to proceed similarly when $\mu_A(x) \neq \mu_A(y)$, but I didn't quit get where I wanted. I think manipulating the convex combination in the right way might be the correct thing. Any hints?
If $t>\mu_A(x)$ and $s>\mu_A(y)$, then $x/t,y/s\in A$ and $$\frac{x+y}{s+t}=\frac{t}{s+t}\frac{x}{t}+\frac{s}{s+t}\frac{y}{s}\in A,$$ so $\mu_A(x+y)\leq t+s$. Now put some control on how much larger $t$ and $s$ are than $\mu_A(x)$ and $\mu_A(y)$, respectively (i.e., $t-\mu_A(x)<\varepsilon$ and $s-\mu_A(y)<\varepsilon$ for some $\varepsilon>0$).