In $\Delta PQR$,$S$ is any point on the side $QR$. Prove that $PQ + QR + RP > 2PS$.
(N/A) In $\Delta PQS$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
$PQ + QS > PS$ ---$(1)$
In $\Delta PRS$,by the triangle inequality theorem:
$PR + RS > PS$ ---$(2)$
Adding equations $(1)$ and $(2)$:
$(PQ + QS) + (PR + RS) > PS + PS$
$PQ + PR + (QS + RS) > 2PS$
Since $S$ is a point on $QR$,$QS + RS = QR$.
Therefore,$PQ + PR + QR > 2PS$.
$PQ + QS > PS$ ---$(1)$
In $\Delta PRS$,by the triangle inequality theorem:
$PR + RS > PS$ ---$(2)$
Adding equations $(1)$ and $(2)$:
$(PQ + QS) + (PR + RS) > PS + PS$
$PQ + PR + (QS + RS) > 2PS$
Since $S$ is a point on $QR$,$QS + RS = QR$.
Therefore,$PQ + PR + QR > 2PS$.
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