Given: S is any point on the side QR

To prove: PQ+QR+RP>2PS.

Proof:

Since in a triangle, sum of any two sides is always greater than the third side.

So in ∆PQS, we have,

PQ + QS > PS …(1)

Similarly, ∆PSR, we have,

PR + SR > PS …(2)

Adding 1 and 2

PQ + QS + PR + SR > 2PS

PQ + PR + QR > 2PS …as PR = QS +SR