Proving a inequality of a point inside a triangle Prove that for any $P$ inside $\triangle ABC$ we have $$AP+CP+BP>1/2(AB+BC+AC)$$
My attempt: Since $P$ is obviously the center point I constructed 3 segments that lead to the midopoints of $AB$ $BC$ and $AC$, $(R,S,T)$ respectively. My argument was $0.5AB+RP>AP$ since $AP>RP>0.5AB$ and the same argument holds for the rest of the midpoints. Then if I removed $RP,SP,TP$ then $AP> 1/2AB$ and so on... is this the correct way to go about this proof? 

By the Triangle Inequality, we have:
$AP+BP\gt AB$;
$AP+CP\gt AC$;
$BP+CP\gt BC$.
Add up, divide by $2$.
Remark: The point $P$ is not necessarily the centre point, it is any point inside the triangle. Parenthetically, there are many notions of centre point of a triangle, which agree for equilateral triangles, but not necessarily for other triangles. 
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