Prove or disprove: $1+\frac{\ln^c(ax)-\ln^ca}{\ln^c(2a)-\ln^ca}-x\leq 0$, where $12$, and $c$ is chosen so that $f'(2)=0$ 
Define:
$$f(x)=1+\frac{\ln^{c}\left(ax\right)-\ln^{c}\left(a\right)}{\ln^{c}\left(2a\right)-\ln^{c}\left(a\right)}-x$$
where  $1< x$ and $a>2$. Assume further that the parameter $c$ is chosen so that $f'(2)=0$. The derivation of $c$ involves the Lambert's function.
Claim :
$$f(x)\leq 0$$

My attempt:
We have :
$$f'(x)=\frac{c(\ln(ax))^{c-1}}{x\left(\ln^{c}\left(2a\right)-\ln^{c}\left(a\right)\right)}-1$$
We substitute $x=\frac{1}{y^{c-1}a}$
The inequality have the form :
$$\ln(u)u=p$$
Wich is just the Lambert's function .See the solution in this link . I cannot proceed further .
How to (dis)prove the first inequality ?
Thanks in advance
 A: Sketch of a proof:
From $f'(2) = 0$, we have
$$\left(\frac{\ln a}{\ln (2a)}\right)^c
= 1 - \frac{c}{2\ln (2a)}.$$
Let
$$g(c) = \left(\frac{\ln a}{\ln (2a)}\right)^c
- 1 + \frac{c}{2\ln (2a)}.$$
We have
\begin{align*}
 g'(c) &= \left(\frac{\ln a}{\ln (2a)}\right)^c\ln \frac{\ln a}{\ln (2a)}
 + \frac{1}{2\ln (2a)},\\
 g''(c) &= \left(\frac{\ln a}{\ln (2a)}\right)^c 
 \left(\ln \frac{\ln a}{\ln (2a)}\right)^2 > 0.
\end{align*}
Clearly $g(0) = 0$, $g'(0) < 0$, $g(1) = \frac{1 - 2\ln 2}{2\ln (2a)} < 0$ and $g(\infty) = \infty$.
Fact 1: $g(1 + \ln(2a))  > 0$.
Fact 2: $1 < c < 1 + \ln (2a)$.

 
Now, we have
$$f''(x) = \frac{c\ln^c (ax)}{[\ln^c (2a) - \ln^c a]x^2 \ln^2 (ax)}[c - 1 - \ln (ax)].$$
Let $x_0 = \frac{1}{a}\mathrm{e}^{c - 1}$. We have $f''(x_0)= 0$, 
$f''(x) > 0$ on $(0, x_0)$, and $f''(x) < 0$ on $(x_0, \infty)$.
Also, by Fact 2, we have $x_0 \in (0, 2)$.
Since $f(2) = 0$ and $f'(2) = 0$, noting that $f''(x) \le 0$ on $[x_0, \infty)$, we have
$f(x) \le f(2) = 0$ for all $x \in [x_0, \infty)$.
Since $f(1) = 0$ and $f(x_0) \le 0$, noting that $f''(x) \ge 0$ on $(0, x_0]$,
we have $f(x) \le \max\{f(1), f(x_0)\} = 0$ for all $x \in (1, x_0]$.
(Note: If $g(x)$ is a convex function on $[a, b]$, then $g(x) \le \max\{g(a), g(b)\}$.)
We are done.
