Multiplying inequalities issues

Solution 1:

As noted in the comments, the second inequality puts looser bounds on $x/\ln x$, so if the first inequality is true, the second will be automatically true as well. Hence there is no contradiction between the two answers.

On the other hand, you might still wonder why the second method is less "precise" than the first. This is because in the inequalities $e \leq x \leq e^2$ and $\frac{1}{2}\leq \frac{1}{\ln x} \leq 1$, the lower (and upper) bounds are not achieved at the same value of $x$. In other words, when $x = e$, you reach the minimum value of $x$, but obtain the maximum value of $1$ for $\frac{1}{\ln x}$. This means that you can't actually reach the minimum $e/2$ that the second method provides.