To solve this problem, we use the triangle inequality that states any side of a triangle is always shorter than the sum of the other two sides. So that, in triangle PQR, QR<PQ+PR ··· Ex.1
Draw a circle centered at T with radius TR, and mark a point S at anywhere on the circle, so ST=TR ··· Ex.2
Cond.1 where points S,Q and T form a triangle SQT. From the triangle inequality, QS<QT+ST. From Ex.2, QT+ST=QT+TR=QR, so that QS<QR. Thus, with Ex.1, we have: QS<PQ+PR
Cond.2 where point S is anywhere on a ray drawn from T thru Q, so QS=|ST-QT|. From Ex.2, |ST-QT|=|TR-QT|, and |TR-QT|<QR, so that QS<QR. Thus, with Ex.1, we have: QS<PQ+PR
Therefore, from cond.1 and cond.2, we have: PQ+PR>QS