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

#### Solution

Let's project $QS$ to intresect $PR$ at $T$
From
$\triangle PQT$, we have
$PQ+PT>QT$

$PQ+PT>SQ+ST$          ----------(1)
$ST+TR > SR$                     ----------(2)

$PQ+PT+ST+TR>SQ+ST+SR$
$PQ+PT+TR>SQ+SR$
$PQ+PR>SQ+SR$
$\Rightarrow SQ+SR<PQ+PR$