Let's prove that $AB'+BC'>CA'$. Then the other two inequalities will be true in the same way, and from these segments it is possible to make a triangle by the inequality of the triangle. In fact, to do this, it is enough to make a rotation centered in $A$, translating $C$ into $B$ and mark the image of the point $B'$ (let it be $B''$), then use the condition that the angles are greater than $120^{\circ}$, from where $C'B''>AC'$ and then apply the inequality the triangle for $B''BC'$.