The problem I'm working on is as follows:
Let $\varphi : [0, \infty) \to [0, \infty)$ have the following properties: Assume \begin{align*} \varphi(0) & = 0 , & \varphi(s) < \varphi (s + t) \leq \varphi (s) + \varphi(t) & \textrm{ for } s \geq 0 , t > 0 \end{align*} Prove that if $d(x, y)$ is symmetric, and satisfies the triangle inequality, then so does $$\delta(x, y) = \varphi (d(x, y)) .$$
I assume I'm supposed to assume that $d$ takes values in $[0, \infty)$.
I've fiddled with this for a few days, and I've yet to figure out which $s, t$ I'm supposed to choose to show that $\delta$ satisfies the triangle inequality. I'm simply at a loss. I know that it just comes down to making the right choice of $s, t$ and doing some (I assume) straightforward algebraic massaging, but like I've said, I've come up empty.
The idea I had for a while was that if I had some $x, y, z$, and I wanted to show that $\delta(x, y) \leq \delta(x, z) + \delta(z, y)$, then I might use \begin{align*} s & = d(x, y) - d(x, z) - d(z, y) \geq 0 , \\ t & = d(x, z) + d(z, y) \end{align*} but I haven't been able to turn this into what I need. Please point me to an appropriate choice of $s, t$.
Hint: In addition to the subadditive property, note that $\varphi$ is also nondecreasing and nonnegative.
Hint: you haven't yet made use of the fact that $d$ satisfies the triangle inequality.
Hint: