$S$ is any point on side $QR$ of a $\triangle PQR$. Show that: $PQ + QR + RP > 2 \, PS$.

(N/A) Given: $A$ point $S$ on side $QR$ of $\triangle PQR$.
To prove: $PQ + QR + RP > 2 \, PS$.
Proof: In $\triangle PQS$,we have:
$PQ + QS > PS$ ..... $(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Now,in $\triangle PSR$,we have:
$RS + RP > PS$ ..... $(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$PQ + QS + RS + RP > 2 \, PS$
Since $QS + RS = QR$,we have:
$PQ + QR + RP > 2 \, PS$
Hence,proved.