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(A) By the power of a point theorem for the circle,we have $PS 	imes ST = QS 	imes SR$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequali

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$B, D$

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(A) By the power of a point theorem for the circle,we have $PS 	imes ST = QS 	imes SR$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for the terms $\frac{1}{PS}$ and $\frac{1}{ST}$,we have:$\frac{\frac{1}{PS}+\frac{1}{ST}}{2} > \sqrt{\frac{1}{PS} 	imes \frac{1}{ST}}$$\Rightarrow \frac{1}{PS}+\frac{1}{ST} > \frac{2}{\sqrt{PS 	imes ST}} = \frac{2}{\sqrt{QS 	imes SR}}$.This proves that option $(B)$ is correct.Next,by the $AM$-$GM$ inequality for $QS$ and $SR$,we have:$\frac{QS+SR}{2} > \sqrt{QS 	imes SR}$$\Rightarrow \frac{QR}{2} > \sqrt{QS 	imes SR}$$\Rightarrow \frac{1}{\sqrt{QS 	imes SR}} > \frac{2}{QR}$.Substituting this into the previous inequality:$\frac{1}{PS}+\frac{1}{ST} > \frac{2}{\sqrt{QS 	imes SR}} > \frac{2 	imes 2}{QR} = \frac{4}{QR}$.This proves that option $(D)$ is correct.Thus,the correct options are $(B)$ and $(D)$.

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