Draw a semicircle whose radius is $1$, and let $A$ and $B$ be the two ends of the arc. Take two points $C$ and $D$ on arc $AB$ so that $A$, $C$, $D$ and $B$ are on the arc in this order.
Let $2a = AC$, $2b = AD$, $\angle CAB = \alpha$, $\angle DAB = \beta$. Then $a = \cos \alpha$, $b = \cos \beta$ and area of the triangle $ACD$ is
$$\begin{aligned} \frac{1}{2}(2a)(2b) \sin(\alpha - \beta) &= 2ab(\sin \alpha \cos \beta - \cos \alpha \sin \beta) \\ &= 2ab(b\sqrt{1-a^2} - a\sqrt{1-b^2}) = 2f(a, b). \end{aligned}$$
Therefore, by taking points $A_1, A_2, \dots, A_n$ on the arc $AB$ with $AA_i = 2a_i$ and considering the sum of the area of triangles $AA_iA_{i+1}$, we conclude that the left side of the given inequality does not exceed $\pi/4$, the half of the area of the semicircle.

Next, we shall prove that $c$ cannot be smaller than $\pi/4$. Take any positive integer $n$. Take points $A = A_0, A_1, A_2, \dots, A_{2^n-1}, A_{2^n} = B$ on the arc $AB$ in this order so that these points divide the arc equally. Let $AA_i = 2a_i$ ($i = 1, 2, \dots, 2^n$). Denote by $S_n$ the part of the semicircle not covered by polygon $A_0A_1 \cdots A_{2^n}$. $S_n$ consists of $2^n$ congruent figures (a fan minus a triangle). Denote each of them by $P_n$. We can easily show that $2|P_{n+1}| < \frac{1}{2}|P_n|$, so $|S_{n+1}| < \frac{1}{2}|S_n|$, hence $|S_n| < \frac{1}{2^{n-1}}|S_1|$. This inequality shows that, if $c$ is less than $\pi/4$, we can take $n$ large enough so that the area of the part which is not covered by polygon $A_0A_1 \cdots A_{2^n}$ be less than $\pi/4 - c$, which yields $c$ inappropriate. Therefore the minimum value of $c$ is $\pi/4$.