If a is the part of the total amount of work that $A$ gets done per day (i.e. $1/6=0.1666\dots$) and b is the part of the work that $B$ gets done per day (i.e. $1/10=0.1$), then $a+b$ is the amount of work that gets done every two days.
No matter who starts, the same amount of work will be done at the end of every two days, and the total amount of work done will look the same for every two-day cycle as well so we need only look at the two first days to see how the accumulated amount of work will progress.
In the interval between every two days, the accumulated amount of work done depends on who starts. If $f(t)$ is a function describing the work done when $A$ works first - then $B$, and $g(t)$ is a function describing the work done when $B$ starts first - then $A$ ($t$ being the number of days), we can for instance write them like this:
f (t) &=
a \cdot t, & t \in [0,1] \\
a + b, \cdot t & t \in (1,2]
g (t) & =
b \cdot t, & t \in [0,1] \\
b + a \cdot t, & t \in (1,2]
($t \in [0,1]$ meaning "for all times between $0$ and $1$", and likewise $t \in (1,2]$ "for all times between $1$ and $2$".)
By inspection, we see that as expected $g(0) = f(0)$, and $g(2) = f(2)$. However, in the interval between $0$ and $2$, the function beginning with the fastest worker (i.e. $f$, since $A$ is fastest) will always be greater than the other. So for any given $t$ that isn't a multiple of $2$, more work will have been done if the fastest worker works the first day.