Let $\phi : [0,\infty) \rightarrow [0,\infty)$ have the following properties: Assume
$$\phi(0)=0, \phi(s)<\phi(s+t)\leq \phi(s)+\phi(t)$$ with $ s\geq0,t>0$.
Prove that if $d(x,y)$ is symmetric and satisfies the triangle inequality, so does $$\delta(x,y) = \phi(d(x,y))$$
I have been reasoning with this for a while, and I feel like it must be a very simple, short solution, but I don't know how to complete it.
I want to show that $\delta(x,y) \leq \delta(x,z) + \delta(y,z)$, or namely that $\phi(d(x,y)) \leq \phi(d(x,z)) + \phi(d(y,z))$.
I know that I have to utilize the triangle inequality of $d(x,y)$ somehow on the left-hand side, working with $\phi(d(x,y))$.
Any help would be appreciated.
Start with $\phi(s)<\phi(s+t)$. Let $s'=d(x,y)\geq 0$ and let $t'=d(x,z)+d(y,z)-d(x,y)\geq 0$. We have :
$$\phi(d(x,y))=\phi(s')<\phi(s'+t')=\phi(d(x,z)+d(y,z))$$
Now, since $\phi(s+t)\leq \phi(s)+\phi(t)$, we directly have :
$$\phi(d(x,z)+d(y,z))\leq \phi(d(x,z))+\phi(d(y,z))$$