Q. AD,BE,CF are the altitudes of an acute angled triangle ABC.
Prove that 1/AD + 1/BE > 1/CF.
Don't you know how to multiply both sides of an inequality by the same number? In general, if x > y and a > 0, then ax > ay. Here I am suggesting multiplying both sides by 2S where S is the area of the triangle ABC and using the fact that S = 1/2 AD * BC, and similarly for other altitudes.