Question

If $S$ is any point in the interior of $\Delta PQR$, prove that $(SQ+SR)<(PQ+PR)$.

Answer 1

Let's project $QS$ to intresect $PR$ at $T$
From $△PQT$, we have
$PQ+PT>QT$
$PQ+PT>SQ+ST$ ----------(1)
$ST+TR>SR$ ----------(2)

Adding equation(1) and (2),
$PQ+PT+ST+TR>SQ+ST+SR$
$PQ+PT+TR>SQ+SR$
$PQ+PR>SQ+SR$
$\Rightarrow SQ+SR<PQ+PR$

Hence proved.