Geometry problem (triangle) How to solve the following problem

Let $P$ be a point inside or outside (but not on) of a triangle $\Delta ABC$. Prove that $PA +PB +PC$ is greater than half of the perimeter of the triangle.

That is, show
$$
PA+PB+PC > 1/2(A+B+C)
$$

Just use the fact that the sum of two sides of a triangle is always bigger then the third side. For example, if $P$ was inside then you look at the (small) triangle $\Delta PBC$, there 
$$PB+PC \geq BC$$
Similarly we get
$$PA+PC \geq AC$$
and
$$PB+PA \geq AB.$$
Now add all the inequalities.
