Inequality involving periodic functions

Question

Let $f,g:\mathbb{R} \to (0,+\infty)$ be two continuous $\tau$-periodic functions. Does there exist a $t \in \mathbb{R}$ so that $$\int_{t-\tau}^{t}\frac{g(t)}{g(s)}f(s)\exp(s-t)ds \geq \overline{f}[1-\exp(-\tau)]$$ where $$\overline{f} = \frac{1}{\tau}\int_0^{\tau}f(s)ds\,?$$ I think one can answer affirmatively when either $f$ or $g$ are constant (notice that the inequality is saturated for all $t$ when both $f$ and $g$ are constant), but the general case eludes me.