Lambert W function style inequality Suppose that for real $x,y \ge 1$ we have the inequality $x/\log x \le y/\log 2$.
Then the inequality
$$
x \le (2/\log 2)y\log y + c
$$
appears to hold for some constant $c$.

What is the smallest such $c$?

I would also accept an argument that no such constant exists.
Depending on how I pose this to Wolfram Alpha, it returns different answers: it seems to sometimes use the wrong branch of the Lambert W function.

$x/\log(x)$  for $x > 1$ has a minimum value of $e$ at $x=e$,  decreasing for $1 < x < e$ and increasing for $e < x < \infty$.  Thus if $y < e \log(2)$ there are no solutions to your inequality; if $y \ge e \log(2)$ the solutions form an interval 
$$ - \frac{y W_{0}(-\log(2)/y)}{\log(2)} \le x \le  - \frac{y W_{-1}(-\log(2)/y)}{\log(2)} $$
where $W_{0}$ and $W_{-1}$ are the principal and "$-1$" branches of the Lambert W function.
So your claim is that
$$- \frac{y W_{-1}(-\log(2)/y)}{\log(2)} < \frac{2 y \log(y)}{\log(2)} + c $$
With $s = \log(2)/y$, this says 
$$ - W_{-1}(-s) < 2 \ln(\ln(2)/s) + c s $$
Note that both $-W_{-1}(-s)$ and $2 \ln(\ln(2)/s)$ are decreasing on $(0, 1/e)$ with values in $(1,\infty)$, and $x \exp(-x)$ is decreasing on  $(1,\infty)$, so this is equivalent to 
$$ -W_{-1}(-s) \exp\left(W_{-1}(-s)\right) > ( 2 \ln(\ln(2)/s) + c s ) \exp\left(- 2 \ln(\ln(2)/s) - c s \right) $$
which simplifies to
$$ s > \frac{s^2 e^{-cs}}{\log(2)^2} \left(2 \log(\log(2)) - 2 \log(s) + c s\right)$$
That is certainly true for sufficiently small $s$, since the right side is 
$O(s^2 |\log(s)|)$ and thus $o(s)$ as $s \to 0+$.  We can adjust $c$ to make it true on the rest of the interval.  
The least possible $c$ is the maximum of 
$$ F(s) = \frac{-2 \log(\log(2)/s) + W_{-1}(-s)}{s} $$
for $0 < s < 1/e$. Numerically it is approximately 
$.455389746$,  occurring at $s \approx .2530875517$. 
