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A straight line through the vertex P 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
$$\frac{1}{PS} + \frac{1}{ST} < \frac{2}{\sqrt{QS+SR}}$$
$$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS+SR}}$$
$$\frac{1}{PS} + \frac{1}{ST} < \frac{4}{QR}$$
$$\frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$$
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