A straight line through the centroid $G$ of a triangle $PQR$ intersects the side $QR$ at the point $S$ and the circumcircle of the triangle $PQR$ at the point $T$. If $S$ is not the centre of the circumcircle, then

$\text{Given: } G \text{ is the centroid of } \triangle PQR$ $\text{Line through } G \text{ meets } QR \text{ at } S \text{ and circumcircle at } T$ $S \neq \text{ centre of the circumcircle}$ $\text{To find: compare } \dfrac{1}{PS} + \dfrac{1}{ST} \text{ with } \dfrac{2}{\sqrt{QS \cdot SR}} \text{ and } \dfrac{4}{QR}$ $\text{Centroid } G: \quad PG : GM = 2 : 1$ $1.\ \text{power of a point at } S$ $2.\ \text{compare the two bounds (AM-GM)}$ $3.\ \text{use the centroid property}$ $ST \cdot SU = QS \cdot SR$ $QS + SR = QR \ \ge \ 2\sqrt{QS \cdot SR}$ $\Rightarrow \ \dfrac{4}{QR} \ \le \ \dfrac{2}{\sqrt{QS \cdot SR}}$ $MP \cdot MT = MQ \cdot MR = \left(\dfrac{QR}{2}\right)^{2}$ $MT = \dfrac{QR^{2}}{4\,MP}$ $\dfrac{1}{MP} + \dfrac{1}{MT} = \dfrac{1}{MP} + \dfrac{4\,MP}{QR^{2}} \ \ge \ \dfrac{4}{QR}$ $\text{Equality} \iff MP = \dfrac{QR}{2} \iff M \text{ is the centre}$ $\dfrac{1}{PS} + \dfrac{1}{ST} > \dfrac{4}{QR}$ $\boxed{\text{Answer: (4)} \quad \dfrac{1}{PS} + \dfrac{1}{ST} > \dfrac{4}{QR}}$