IIT JEE Advanced 2008 Paper 1

Since PT and QR are two secant of circumcircle of triangle PQR, we get
PS × ST = QS × SR

Now, using A.M. – G.M. inequality, we get

1 PS + 1 ST

2

> √

1

PS

×

1

ST

⇒

1

PS

+

1

ST

> 2√

1

QS×SR

⇒

1

PS

+

1

ST

>

2

√QR×SR

(1)
Now,

QS+SR

2

≥ √QS×SR
⇒

QR

2

≥ √QS×SR
⇒

1

√QS×SR

>

2

QR

⇒

2

√QS×SR

>

4

QR

(2)
From Eqs. (1) and (2), we get

1

PS

+

1

ST

>

4

√QR