$S$ is any point in the interior of $\triangle PQR$. Show that $SQ + SR < PQ + PR$.
(N/A) Produce $QS$ to intersect $PR$ at $T$.
From $\triangle PQT$,we have:
$PQ + PT > QT$ (The sum of any two sides of a triangle is greater than the third side)
$PQ + PT > SQ + ST$ ..... $(1)$
From $\triangle TSR$,we have:
$ST + TR > SR$ ..... $(2)$
Adding $(1)$ and $(2)$,we get:
$PQ + PT + ST + TR > SQ + ST + SR$
$PQ + PT + TR > SQ + SR$
Since $PT + TR = PR$,we have:
$PQ + PR > SQ + SR$
Therefore,$SQ + SR < PQ + PR$.
From $\triangle PQT$,we have:
$PQ + PT > QT$ (The sum of any two sides of a triangle is greater than the third side)
$PQ + PT > SQ + ST$ ..... $(1)$
From $\triangle TSR$,we have:
$ST + TR > SR$ ..... $(2)$
Adding $(1)$ and $(2)$,we get:
$PQ + PT + ST + TR > SQ + ST + SR$
$PQ + PT + TR > SQ + SR$
Since $PT + TR = PR$,we have:
$PQ + PR > SQ + SR$
Therefore,$SQ + SR < PQ + PR$.
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