$d(0,\alpha x+\beta y) \leq |\alpha|d(0,x)+|\beta|d(0,y);\ |\alpha|+|\beta|\leq 1$

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Let $X$ be a metrizable topological vector space with the metric $d$ and $\alpha, \beta$ be scalars (complex or real). Such a metric $d$ has the following properties: $$d(x+z,y+z)=d(x,y),\ d(0,\alpha x) \neq |\alpha|d(0,x),\ \forall x,y,z \in X$$ I am trying to prove or disprove the following but I couldn't: $$d(0,\alpha x+\beta y) \leq |\alpha|d(0,x)+|\beta|d(0,y);\ |\alpha|+|\beta|\leq 1,\ x,y \in X$$

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d(0,x+y) = d(-x,y) <= d(-x,0) + d(0,y) = d(0,x) + d(0,y)

d(0,ax) = |a|d(0,x)