Cesenatico

Solution:

Let us call $x = MA^{\prime}$, $y = MB^{\prime}$, $r = CA^{\prime}$, $s = CB^{\prime}$. It is clear that
$$A^{\prime}B^{\prime} = x - y = s - r.$$
Let us denote by $A_{2}$ and $A_{3}$ the points of tangency of $AB$ and $AC$ with $\alpha$, and by $B_{2}$ and $B_{3}$ the points of tangency of $AB$ and $BC$ with $\beta$.

Now let us repeatedly use the fact that, drawing the tangents from a point $X$ external to a circle $\gamma$, and calling $M$ and $N$ the points of tangency, we have $XM = XN$.
We have that $x = MA_{2}$, $y = MB_{2}$, $r = CA_{3}$, $s = CB^{\prime}$. Let us then set
$$t = AA_{2} = AA_{3}, \quad u = BB_{2} = BB_{3}.$$
Since $M$ is the midpoint of $AB$, we have $x + t = y + u$. On the other hand, $t = b - r$, $u = a - s$, from which $x + b - r = y + a - s$ and therefore
$$(x - y) + (s - r) = a - b.$$
Since $a - b > 0$, in (1) the $+$ sign holds and the claim is proved.