I'm trying to solve the following question:
assume $f : [0, \infty) \rightarrow \mathbb{R}$ is convex and continuously differentiable where $f(0)=0$ and $f'(0)=0$. $X$ is a normed vector space on $\mathbb{R}$ and $dim X \geq 2$. Show that for each $0 \neq x \in X$ and real $\alpha > 0$ there exist $y \in X$ such that $||y||=\alpha$ and $f(||x+y||)=f(||x||)+f(||y||)$
My attemp: defining $y=\frac{\alpha}{||x||}$ it's obvious that $y \in X$ and $||y||=\alpha$.
for the equality given the fact that $f(0)=0$ we can conclude that $f(\lambda x) \leq \lambda f(x)$ if $0 \leq \lambda \leq 1$ and from that it follows that $f(||x||) + f(||y||) \leq f(||x||+||y||)$
for the other direction I'm not sure what to do. Since $f'(0) =0$ and $f$ is convex then f is non-decreasing. I tried to use this fact with triangle inequality of the norm but I wasn't able to reach the inequality.