Question

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

Answer 1

In ∆PQS, 

PQ + QS > PS …..(i) [Sum of any two sides of a triangle is greater than the third side] 

Similarly, in ∆PSR, 

PR + SR > PS …(ii) [Sum of any two sides of a triangle is greater than the third side] 

∴ PQ + QS + PR + SR > PS + PS 

∴ PQ + QS + SR + PR > 2PS 

∴ PQ + QR + PR > 2PS [Q-S-R]