For which values of $a$ does $d\ge ac\ln c\implies d\ge c\ln d$? For which values of $a>0$ is it true that for all $c,d>0$, 
$\hspace{.2 in}d\ge ac\ln c\implies d\ge c\ln d$?
I believe that this is true for $a\ge2$, (see Showing if $n \ge 2c\log(c)$ then $n\ge c\log(n)$),
and that it is not true for $a=1$, since $4\ge3\ln3$, but $4<3\ln 4$.
(I think it is also enough to consider $c>e$, since $0<c\le e\implies d\ge c\ln d$ for any $d>0$.)
 A: Note that if the statement holds for some $a=a_0$, then it holds for all $a>a_0$ as well. As OP notes, the statement is false for $a=1$, so it is false for $a<1$ as well. As such, consider $a>1$. We are trying to determine if we have
$$\inf\limits_{d\ge ac\ln c, d>1}{\frac{d}{\ln d}}\ge c $$
for any $c>0$, or in other words whether or not
$$\inf\limits_{c>0}\left(\frac{1}{c}\inf\limits_{d\ge ac\ln c, d>1}{\frac{d}{\ln d}}\right) \ge 1. $$
(Clearly we don't have to consider $d\le 1$.)
Notice that $d\mapsto\frac{d}{\ln d}$ has a minimum at $d=e$, with $\frac{e}{\ln e} = e$, and it is increasing for $d>e$. As such, we only need to consider $c\ge e$. For $a>1$ and $c\ge e$, we have
$$d\ge ac\ln c \implies d\ge ae\ln e > e.$$
As such, the minimum of $\frac{d}{\ln d}$ for $d\ge ac\ln c$ occurs at $d = ac\ln c$, so
$$\inf\limits_{d\ge ac\ln c}{\frac{d}{\ln d}} = \frac{ac\ln c}{\ln(ac\ln c)}. $$
As such, we ask whether or not
$$\inf\limits_{c\ge e}\left(\frac{a\ln c}{\ln(ac\ln c)}\right)\ge 1.$$
Since $\ln(ac\ln c)>0$ for $a>1$ and $c\ge e$, we have
$$\exists c\ge e: \frac{a\ln c}{\ln(ac\ln c)}< 1 \\
\iff \exists c\ge e: a\ln c < \ln(ac\ln c) = \ln a + \ln c + \ln\ln c \\
\iff \exists c\ge e: (a-1)\ln c - \ln\ln c - \ln a < 0 \\
\iff \inf\limits_{c\ge e}{(a-1)\ln c - \ln\ln c} - \ln a < 0.$$
Let $f(c) = (a-1)\ln c - \ln\ln c - \ln a$. We then have
$ f'(c) = \frac{a-1}{c} - \frac{1}{c\ln c} = \frac{1}{c}\left(a-1-\frac{1}{\ln c}\right) $, which implies that $f'(c) = 0$ at $c = e^{1/(a-1)}$, $f'(c)<0$ for $c< e^{1/(a-1)}$, and $f'(c)>0$ for $c > e^{1/(a-1)}$. Thus, $f$ attains its minimum at $c = e^{1/(a-1)}$, with
$$ f(e^{1/(a-1)}) = (a-1)\left(\frac{1}{a-1}\right) - \ln\left(\frac{1}{a-1}\right) - \ln a = 1 - \ln\left(\frac{a}{a-1}\right).$$
This is negative precisely when $\frac{a}{a-1} > e$, i.e. when $a < \frac{e}{e-1}$. Notice that $e^{1/(a-1)}\ge e$ iff $a\le 2$, and $\frac{e}{e-1}<2$. Thus, there exists $c\ge e$ such that $\frac{a\ln c}{\ln(ac\ln c)}<1$ precisely when $a < \frac{e}{e-1}$, i.e. $\inf\limits_{c\ge e}{\left(\frac{a\ln c}{\ln(ac\ln c)}\right)}<1$ precisely when $a < \frac{e}{e-1}$. 
Thus, the desired statement is true precisely when $a\ge \frac{e}{e-1}$.
A: 1) First we show by contradiction that the statement is true for $\displaystyle a\ge\frac{e}{e-1}$: 
$\hspace{.2 in}$Let $d\ge ac\ln c$,  and assume that $d<c\ln d$.
$\hspace{.2 in}$Then $c\ln d>d\ge ac\ln c\implies\ln d>a\ln c=\ln c^a\implies d>c^a$, 
$\hspace{.2 in}$so $ c\ln d>d>c^a\implies \ln d>c^{a-1}>\left(\frac{d}{\ln d}\right)^{a-1}\implies(\ln d)^a>d^{a-1}\implies \ln d>d^{1-\frac{1}{a}}$.
$\hspace{.2 in}$This gives a contradiction, since $1-\frac{1}{a}\ge\frac{1}{e}$ and $\ln x\le x^k$ for all $x>0$ if $k\ge\frac{1}{e};\;$ so $d\ge c\ln d$.
2) If $1<a<\frac{e}{e-1},\;\;$ let $c=e^{1/(a-1)}$ and let $d=ac\ln c,\;$ so $d=\frac{a}{a-1}e^{1/(a-1)}$.
$\hspace{.2 in}$Then $d<c\ln d\iff ac\ln c<c\ln d\iff a\ln c<\ln d\iff c^a<d\iff$
$\hspace{.2 in} e^{a/(a-1)}<\frac{a}{a-1}e^{1/(a-1)}\iff \frac{a-1}{a}<e^{-1}\iff 1-\frac{1}{a}<\frac{1}{e}\iff a<\frac{e}{e-1},\;$ so $d<c\ln d$.
Therefore the statement is valid exactly when $a\ge\frac{e}{e-1}$.
