A non-polynomial Young function satisfying a power-like condition This post asked, essentially, for an example of a "non-polynomial" invertible increasing function $f\colon[0,\infty)\to[0,\infty)$ such that $f(0)=0$ and
\begin{equation}
    f(cu)f(t)\le f(tu) \tag{1}\label{1}
\end{equation}
for some real $c>0$, all $u\in[0,1]$, and all real $t\ge0$; the question in that post is a bit rephrased here. An answer to that question was given.
Then the OP said in a comment that the condition that $f$ be a Young function was missing in that post. A simple example of a Young function $f$ satisfying the above conditions will be given in the answer below.
 A: Consider the Young function $f$ defined by
\begin{equation*}
    f(x):=\ln(1 + x) - x + x^2/2. \tag{$*$}\label{*}
\end{equation*}
Take $c=1$ and then rewrite \eqref{1} as
\begin{equation*}
        g(t):=f(u)f(t)-f(tu)\le0. \tag{2}\label{2}
\end{equation*}
We have $g(0)=0$ and
\begin{equation*}
    g'(t)=f(u)f'(t)-uf'(tu)
    =\frac{t^2}{2 (1 + t)}\,g_1(u),
\end{equation*}
where
\begin{equation*}
    g_1(u):=2 \ln(1+u)-\frac{u \left((t+2) u^2+(2 t-1) u+2\right)}{1+t u}. 
\end{equation*}
Next, $g_1(0)=0$ and
\begin{equation*}
    g'_1(u)=-\frac{2 u^2 \left(t^2 u (u+2)+t \left(2 u^2+3 u+3\right)+3 u+2\right)}
    {(1+u) (1+t u)^2},
\end{equation*}
which is manifestly $\le0$ for $u\in[0,1]$ and $t\ge0$. So, $g_1\le0$ and hence $g$ is decreasing. Thus, \eqref{2} follows, and hence \eqref{1} follows as well.

User Guy Fsone has now stated, in a comment to the question above, "I need More examples (2 or 3) of invertible non-polynomial Young functions $f$ satisfying the condition: There exists $c>0$ such that $f(cu)f(t)\le f(tu)$ for all $u\in[0,1]$ and $t\ge1$."
Let us therefore provide an entire series of examples generalizing the example above. To do that, take any natural $m$ (so that $m\ge1$) and let
\begin{equation}
    f(x):=f_m(x):=\ln(1+x)-\sum_{k=1}^{2m}\frac{(-1)^{k-1}}k\,x^k \tag{3}\label{3}
    =\int_0^x\frac{a^{2m}\,da}{1+a}  
\end{equation}
for real $x\ge0$, so that the function $f$ given by \eqref{*} is $f_1$. Moreover, $f=f_m$ is an invertible non-polynomial Young function.
Take $c=1$ and then rewrite \eqref{1} as \eqref{2}, as was done previously.
We have $g(0)=0$ and
\begin{equation*}
    g'(t)=f(u)f'(t)-uf'(tu)
    =\frac{t^{2m}}{1 + t}\,g_1(u),
\end{equation*}
where
\begin{equation*}
    g_1(u):=f(u)-\frac{u^{2m+1}}{1 + ut}\,(1+t). 
\end{equation*}
Next, $g_1(0)=0$ and
\begin{equation*}
    g'_1(u)=-\frac p{(1+u) (1+t u)^2},
\end{equation*}
where
\begin{equation*}
    p:=u^{2 m} \left(u^2 \left((2 m-1) t^2+2 m t\right)+u \left(2 m t^2+(4 m-1) t+2 m+1\right)+(2 m+1) t+2 m\right), 
\end{equation*}
so that $g'_1(u)$ is manifestly $\le0$ for $u\ge0$ and $t\ge0$ (and $m\ge1$). So, $g_1\le0$ and hence $g$ is decreasing, from $g(0)=0$. Thus, \eqref{2} follows, and hence \eqref{1} follows as well.

Yet another family of such examples is given by the formula
\begin{equation*}
    f(x):=f_b(x):=\frac{x^b}{1+x}
\end{equation*}
for real $b>2$. Here all the desired conditions can be checked easily.
More generally, for each real $c>0$, consider the following set of functions (indexed by three parameters $b,p,q$):
\begin{equation*}
F_c:=\big\{f_{c;\,b,p,q}\colon b\in(2,\infty), p\in(0,c^b], q\in(0,c^b]\big\},  
\end{equation*}
where the function $f_{c;\,b,p,q}$ is defined by the formula
\begin{equation*}
    f_{c;a,p,q}(x):=\frac{x^b}{p+qx}
\end{equation*}
for real $x\ge0$. Then it is easy to check that every function $f\in F_c$ is an invertible non-polynomial Young function satisfying condition \eqref{1} for all real $u,t\ge0$.
