Suppose that $$\exp(-a + b - \log(c)) \leq x \leq \exp(a + b -\log(c)).$$
Moreover, suppose that $0 \leq x \leq \frac{1}{e}$. I would like to conclude that $$-x\log(x) \leq \frac{-\exp(-a +b)}{c}(a + b - \log(c)).$$
First, observe that $$x \geq \exp(-a + b - \log(c)) \implies-x \leq -\exp(-a + b - \log(c)).$$
Furthermore, since $\log(x)$ is an increasing function, we have that $$\log(x) \leq a + b -\log(c).$$
If we didn't need to care about signs, we could conclude our result. I think I need to use the fact that $f(x) = -x\log(x)$ is increasing in $(0, 1/e]$ and that $\log(x) < 0$ in $(0, 1/e]$.
I don't think that this is true:
$$-x\log(x) \leq \frac{-\exp(-a +b)}{c}(a + b - \log(c)) \Leftrightarrow $$
$$ x\log(x) \geq \exp(-a +b - \log(c))(a + b - \log(c)) \Leftrightarrow $$
$$ \log(x) \geq \frac{\exp(-a +b- \log(c))}{x}(a + b - \log(c)).$$
On the other hand
$$\exp(-a + b - \log(c)) \leq x \leq \exp(a + b -\log(c))$$ permits choosing $a$, $b$ , $c$ in such a way that $\frac{\exp(-a +b- \log(c))}{x}$ is arbitrarily close to one, and strictly smaller than one. But we have $\log(x) \leq (a + b - \log(c))$ and thus we cannot have $\log(x) \geq 1 \times (a + b - \log(c))$ contemporaneously