Question

In ΔPQR, S is any point on the side QR. Show that PQ + QR + RP $>$ 2 PS

Answer 1

In $△$ PQS, we have

PQ + QS $>$ PS ...(i)

[$\because$ Sum of the two sides of a $△$ is greater than the third side]

Similarly, in $△$PRS, we have

RP + RS $>$ PS

Adding (i) and (ii), we get

(PQ + QS) + (RP + RS) $>$ PS + PS

$\Rightarrow$ PQ +(QS + RS) + RP $>$ 2 PS

$\Rightarrow$ PQ + QR + RP $>$ 2 PS [ $\because$ QS + RS = QR]