Let ABC be an equilateral triangle and let M be a point that does not lie on the circle circumscribed on the triangle. Show that the segments AM, BM and CM can form sides of a triangle. Hint: Check that the triangle inequalities $AM<BM+CM$, $BM<AM+CM$ and $CM<AM+BM$ hold.
My attempt:
The circle circumscribed on the triangle is shown in the figure. At the intersection of BM and AC, we put a point P. Now we find that in $\triangle{AMC}$: $$AP+PM \geq AM$$ and $$PC+PM\geq CM$$ Then we form the equivalent triangle inequalities for $\triangle{ABC}$, which are $$AP+PB\geq AB$$ and $$BP+PC\geq BC.$$ . Furthermore, we can form the additional triangle inequalities $$MC+CB\geq BM$$ and $$BA+AM\geq BM$$. Clearly, we see that every side in each given triangle inequalities are part of triangles, and hence $AM, \ BM, \ CM$ all form sides of a triangle.
But if one wants to prove that $AM, BM, CM$ alone form a triangle, it is not so trivial from the given reasoning. Is there any way to do this here?
Any hints appreciated