I stumbled across the inequality below, which should be correct...I can't come up with a proof (so frustrating :S). Maybe someone can help? Thanks a lot!
For $a\geq 0$, $b\in \mathbb{R}$
$$ab\leq a\log{a} - a + e^b.$$
(sorry for the simplicity... I am a bit out of practice)
One way of proving such inequalities without thinking too much, is by considering the extrema.
Move everything to to the r.h.s. and consider the function $$f(a, b) = a\log a - a + e^b - ab$$ We would like to prove that $$f(a, b) \ge 0,\; (a,b) \in \mathbb{R}_+ \times \mathbb{R}$$ If $a = 0$, then $f(0, b) = e^b$ (by continuity $a\log a = 0$), so in this case clearly $f(0, b) > 0$. From now on we can assume $a > 0$. Let's minimize $f$ for a fixed $a$. Since exponent grows faster then linear function, $f(a,b) \to +\infty$ when $ b \to +\infty$. If $b\to-\infty$, then $e^b - ab \to +\infty$ because $a > 0$, so we also have $f(a,b) \to +\infty$. Thus the minimum of $f$ occurs at an extremum $$\frac{\partial f(a,b)}{\partial b} = e^b - a = 0 \Rightarrow b = \log a$$ Plugging in this point to compute the minimal value for a fixed $a$, we obtain $$f(a, \log a) = a\log a - a + a - a\log a = 0$$ Since this is a minimum, for all other points $b$ we must have $f(a, b) \ge 0$. Since this is true for any $a$ we have proved the inequality.
P.S. If you want to think even less, you can just minimize $f$ directly as a function of two variables. But then you have to prove that the extrema aren't reached on the boundary/at infinity, which seems to be harder for this particular problem.