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Question

In the following figure, S is any point on the side QR of Δ PQR. Prove that PQ + QR + RP > 2 PS.

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Solution

The sum of any two sides of a triangle is grater than the third side.
In PQS and PSR, we have
PQ + QS > PS ..... (i)
PR + SR > PS ..... (ii)
Adding (i) and (ii), we get
PQ + PR + (QS + SR) > 2PS
Since QS + SR = QR, therefore
PQ + PR + QR > 2PS
Hence, PQ + QR + RP > 2PS.

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