S is any point on side QR of a ΔPQR. Show that: PQ + QR + RP > 2 PS.

Given: S is any point on side QR of a ΔPQR

In ΔPQS,

PQ + QS > PS (sum of two sides is greater than the third side) ...(1)

Similarly, In ΔPRS,

SR + RP > PS (sum of two sides is greater than the third side) ...(2)

Add (1) and (2)

PQ + QS + SR + RP > 2 PS

⇒ PQ + (QS + SR) + RP > 2 PS

⇒ PQ + QR + RP > 2 PS (as, QS + SR = QR)

Hence, proved.