Suppose $(M,d)$ is metric. I have proven that if $\psi\colon[0,\infty)\to[0,\infty)$ is non-decreasing, subadditive and satisfies $\psi(x)=0\iff x=0$ for $x\ge0$, then $$\rho(x,y)=\psi(d(x,y))$$ is a metric on $M$.
But I want to `tweak' the restriction of this statement, I want to use differentiability of $\psi$. I found the following:
Suppose $(M,d)$ is metric and $\psi\colon[0,\infty)\to[0,\infty)$ is differentiable with continuous non-increasing derivative $\psi'$ and $\psi(0)=0$. Then $\psi$ is non-decreasing and subadditive.
I have also proven this statement, but is $\rho=\psi(d(x,y))$ a metric in this situation? And if not, what is a sufficient condition on the derivative $\psi'$ to turn $\rho$ into a metric on $M$?