Question

In $\Delta PQR$, $S$ is any point on side $QR$. Show that $PQ+QR+RP>2PS$.

Answer 1

In $\Delta PQR$,
$PQ+QS>PS$ …(i) ( $\because$ the sum of any two sides of a triangle is greater than the third side)
In $\Delta PRS$,
$PR+RS>PS$ ($\because$ The sum of any two sides of a triangle is greater than the third side)
On adding (i) and (ii), we get
$(PQ+QS)+(PR+RS)>2PS$
$PQ+(QS+RS)+PR>2PS$
$PQ+QR+PR>2PS$
$PQ+QR+RP>2PS$
Hence proved.