Question

IN ΔPQR ,S is any point on the side on QR show that PQ+QR+RP > 2PS 

Answer 1

In ΔPQS
PQ+QR>PS   ........(1)
 (Sum of two sides of a Δ is greater than the third side)
PR+RS>PS    .........(2)
 (Sum of two sides of a Δ is greater than the third side)
On adding (1) and (2), we get  :
PQ+QR+PR+RS>2PS
PQ+QR+RP > 2PS

Answer 2

Answer :

$<b><u>Given</u>$: S is any point on side QR of a ΔPQR

In ΔPQS,

$<b>$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)

$<b>Add (1) and (2)</b>$

PQ + QS + SR + RP > 2 PS

⇒ $<b>PQ + (QS + SR) + RP > 2 PS</b>$

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

Hence, proved.