Final Round

Let $ZM$ be parallel to $AC$ (see the figure).

It is evident that $MZCA$ is a parallelogram, so $MZ=AC$ and $MA=CZ$. By condition, $AX=CZ$, so $AX=MA$. Therefore, the altitude $YA$ of the triangle $MYX$ is a median, hence this triangle is an isosceles triangle and $MY=YX$.

Since the length of the segment $MZ$ is less than or equal to the length of the broken line $MYZ$, we obtain $XY+YZ=MY+YZ\ge MZ=AC$, as required.