Given a specific function $f$, how to compute the left-inverse of $f$ in the sense of $\approx$? For a non-negative function $\varphi$ defined on $[0,\infty)$, the left-inverse $\varphi^{-1}$ of $\varphi$ is defined by setting, $\forall t\geq 0$,
$$\varphi^{-1}(t):=\inf\{u\geq0:\varphi(u)\geq t\}.$$
For two functions $\varphi$ and $\psi$, we say that $\varphi\approx\psi$, if there exist constants $c_1,c_2>0$, such that $\forall t\geq0$,
$$c_1\varphi(t)\leq\psi(t)\leq c_2\varphi(t).$$
Let $p\geq1$, define $f(t):=t^p\ln(e+t)$, $t\geq0$ (this is an example appearing in the paper about equations). How to compute the $f^{-1}$ in the sense of $\approx$? I guess its
answer is $\forall t\geq0$, $$f^{-1}(t)\approx\left[\frac{t}{\ln(e+t)}\right]^{1/p}\quad \text{or}\quad f^{-1}(t)\approx \min\left\{t^{1/p},\left[\frac{t}{\ln(e+t)}\right]^{1/p}\right\},$$ but I do not know how to compute $f^{-1}$ in the sense of $\approx$, I need some help.
 A: If $f(t):=t^p\ln(e+t)$ for some real $p>0$ and all real $t\ge0$, then
\begin{equation*}
    f^{-1}(u)\approx g(u):=\frac{u^{1/p}}{\ln^{1/p}(e+u^{1/p})} \label{1}\tag{1}
\end{equation*}
for real $u\ge0$, where the symbol $\approx$ is used in the sense defined in your post.
Indeed, the function $f\colon[0,\infty)\to[0,\infty)$ is continuous and strictly increasing, from $f(0)=0$ to $f(\infty-):=\lim_{t\to\infty}f(t)=\infty$. So, $f^{-1}$ is the true inverse of $f$. That is, for all nonnegative real $t$ and $u$ we have
\begin{equation*}
    f(t)=u\iff f^{-1}(u)=t. \label{2}\tag{2}
\end{equation*}
Next,
\begin{equation*}
    Kh(u)\le g(u)\le u^{1/p}, \label{3}\tag{3}
\end{equation*}
where
\begin{equation*}
h(u):=\min(u^{1/p},u^{1/(2p)})=u^{1/p}\,1(u\le1)+u^{1/(2p)}\,1(u>1) \label{4}\tag{4}
\end{equation*}
and $K\in(0,\infty)$ depends only on $p$ (but not on $u$); here and in what follows, $u$ is an arbitrary nonnegative real number, unless further specified. (The first inequality in \eqref{4} holds because $g(0)=h(0)=0$, the functions $g$ and $h$ are continuous and strictly positive on $(0,\infty)$, and for $r(u):=g(u)/h(u)$ $(u>0)$ we have $r(0+)=1>0$ and $r(\infty-)=\infty>0$.)
By the second inequality in \eqref{3},
\begin{equation*}
f(g(u))=u\,\frac{\ln(e+g(u))}{\ln(e+u^{1/p})}\le u=f(f^{-1}(u)). 
\end{equation*}
Therefore and because $f$ is strictly increasing, we have
\begin{equation*}
g(u)\le f^{-1}(u).\label{5} \tag{5}
\end{equation*}
On the other hand, by the first inequality in \eqref{3}, for real $C\ge1$
\begin{equation*}
    f(Cg(u))=C^pu\,\frac{\ln(e+Cg(u))}{\ln(e+u^{1/p})} \label{6}\tag{6}
    \ge C^pu R(u),
\end{equation*}
where
\begin{equation*}
R(u):=\frac{\ln(e+Kh(u))}{\ln(e+u^{1/p})}. 
\end{equation*}
The function $R$ is continuous and strictly positive on $(0,\infty)$, with $R(0+)=1>0$ and $R(\infty-)=1/2>0$. So, $R(u)\ge\ell$ for all real $u\ge0$, where $\ell\in(0,\infty)$ depends only on $p$ and $K$, and hence only on $p$ (but not on $u$).
Taking now any real $C\ge1$ such that $C^p\ell\ge1$, from \eqref{6} we get $f(Cg(u))\ge u=f(f^{-1}(u))$, which implies $Cg(u)\ge f^{-1}(u)$.
Thus, in view of \eqref{5}, $   g(u)\le f^{-1}(u)\le Cg(u)$. So, \eqref{1} is proved.
A: As a supplement and extension of Iosif's answer, let me mention that, more generally, if $$f(t) \sim t^{pq} (\ell(t^q))^p \qquad \text{as } t \to \infty$$ for a slowly varying function $\ell$, then $$f^{-1}(s) \sim s^{1/(pq)} (\ell^\#(s^{1/p}))^{1/q} \qquad \text{as } s \to \infty,$$ where $\ell^\#$ is another slowly varying function: the de Bruijn conjugate of $\ell$. See Proposition 1.5.15 in Regular Variation by Bingham, Goldie and Teugels. If $\ell(t) = \log t$, then $\ell^\#(s) \sim 1 / \log t$, and so we recover an asymptotic variant of the desired result (with $q = 1$).
A variant with asymptotic equality ($\sim$) replaced by comparability ($\approx$) is essentially given as Problem 1.11.19.
