A straight line through the vertex \(P\) of a \(\triangle P Q R\) intersects the side \(Q R\) at the point \(S\) and the circumcircle of the \(\triangle P Q R\) at the point \(T\). If \(S\) is not the centre of the circumcircle, then

Let a straight line through the vertex \(P\) of a given \(\triangle P Q R\) intersects the side \(Q R\) at the point \(S\) and the circumcircle of \(\triangle P Q R\) at the point \(T\).
Points \(P, Q, R, T\) are concyclic, hence
\[
\begin{array}{rlrl}
P S \cdot S T & =Q S \cdot S R & & \\
\Rightarrow \quad P_1 & \geq 2 \sqrt{P S \cdot S T} & \\
& \text { Now, } \frac{P S+S T}{2} & \geq \sqrt{P S \cdot S T} & {[\because \mathrm{AM} \geq \mathrm{GM}]}
\end{array}
\]


and \(\quad \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}\)
Also, \(\quad \frac{S Q+Q R}{2} \geq \sqrt{S Q \cdot S R}\)
\(\Rightarrow \quad \frac{Q R}{2} \geq \sqrt{S Q \cdot S R}\)

\[
\begin{array}{ll}
\Rightarrow & \frac{1}{\sqrt{S Q \cdot S R}} \geq \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} \geq \frac{4}{Q R} \\
\text { Hence, } & \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{Q S \cdot S R}} \geq \frac{4}{Q R} .
\end{array}
\]