Looking for non-polynomial functions: with the growth condition: $\phi\big(\theta \frac{s}{t}\big) \leq \frac{\phi(s)}{\phi(t)}$ I am for example(s) of an invertible Convex or concave function $\phi: [0,\infty)\to [0, \infty)$ such that $\phi(0)=0$ and there exists $\theta>0$ and for all $s\leq t$ we have
\begin{align}\label{EqI}\tag{I}
\phi\big(\theta \frac{s}{t}\big)    \leq \frac{\phi(s)}{\phi(t)} \qquad\text{or equaly}  \qquad  \theta  \leq \phi^{-1}\big(\frac{s}{t}\big)\frac{\phi^{-1}(t)}{\phi^{-1}(s)}
\end{align}
The most simple class consists of polynomial functions of the form $\phi(t)= ct^p$ with $c>0$ and $p>0$.
Question: Are there other possible non-polynomial examples satisfying $\eqref{EqI}$?
As an attempt with $\phi(t)= e^{t^\alpha}-1$, I wonder if there is a constant $c>0$ such that
$$ \ln(t+1)\ln\left(\frac{s}{t}+1\right)\geq c\ln(s+1),\qquad \text{for all $0\leq s\leq t$}.$$
 A: With $f:=\phi$, $u:=s/t$, and $c:=\theta$, the desired inequality can be rewritten as
$$f(cu)f(t)\le f(tu) \tag{1}$$
for $u\in[0,1]$ and real $t\ge0$.
Let us show that (1) holds with $c=1$ if $f(x)\equiv\ln(1+x)$. That is, we have to show that
$$g(t):=\ln(1+u)\ln(1+t)-\ln(1+tu)\le0$$
for $u\in[0,1]$ and real $t\ge0$. We have $g(0)=0$ and
$$g'(t)=\frac{\ln(1+u)}{1+t}-\frac u{1+tu}
\le \frac u{1+t}-\frac u{1+tu}\le0$$
for $u\in[0,1]$ and real $t\ge0$. So, the desired result follows.

More generally, any concave function $f$ such that $f(0)=0$ and $0<f'\le1/c$ satisfies (1); here $f'$ denotes the right derivative of $f$. Indeed, then $0\le f(cu)\le u$ for all real $u\ge0$ and hence for
$$h(t):=f(cu)f(t)-f(tu)$$
we have $h(0)=0$ and
$$h'(t)=f(cu)f'(t)-uf'(tu)\le uf'(t)-uf'(tu)\le0$$
for $u\in[0,1]$ and real $t\ge0$.
So, taking any positive decreasing function $f_1$ on $[0,\infty)$ and then letting $c:=1/f_1(0)$ and
$$f(x):=\int_0^x f_1(y)\,dy,$$
for real $x\ge0$, one has (1) satisfied.

The inequality
$$ \ln(t+1)\ln\left(\frac{s}{t}+1\right)\ge c\ln(s+1)\qquad \text{for all $0\le s\le t$} \tag{2}$$
in your post does not hold for any real $c>0$ if $s$ is small and $t$ is large, because for such $s$ and $t$ we have $\ln\left(\frac{s}{t}+1\right)\sim\frac{s}{t}$ and $\ln(s+1)\sim s$, whereas $\ln(t+1)/t\to0$ as $t\to\infty$.
