Question

In ∆PQR, if S is any point on the side QR, show that PQ + QR + RP > 2PS.

Answer 1

$$\begin{gathered} PQ+QS>PS...\left(1\right)[\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{its}\mathrm{third}\mathrm{side}] \\ PR+SR>PS...\left(2\right)[\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{its}\mathrm{third}\mathrm{side}] \end{gathered}$$

On adding (1) and (2) we get:
$$\begin{gathered} PQ+QS+PR+SR>2PS \\ \Rightarrow PQ+QR+PR>2PS \end{gathered}$$

Hence, proved.