Solution:
As PR $>P Q$
$\therefore \angle \mathrm{PQR}>\angle \mathrm{PRQ}$ (Angle opposite to larger side is larger) ... (1)
$P S$ is the bisector of $\angle Q P R$.
$\therefore \angle Q P S=\angle R P S \ldots(2)$
$\angle P S R$ is the exterior angle of $\triangle P Q S$.
$\therefore \angle \mathrm{PSR}=\angle \mathrm{PQR}+\angle \mathrm{QPS} \ldots(3)$
$\angle P S Q$ is the exterior angle of $\triangle P R S$.
$\therefore \angle \mathrm{PSQ}=\angle \mathrm{PRQ}+\angle \mathrm{RPS} \ldots(4)$
Adding equations $(1)$ and $(2)$, we obtain
$\angle \mathrm{PQR}+\angle \mathrm{QPS}>\angle \mathrm{PRQ}+\angle \mathrm{RPS}$
$\Rightarrow \angle \mathrm{PSR}>\angle \mathrm{PSQ}[U$ sing the values of equations $(3)$ and $(4)]$
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