Q. AD,BE,CF are the altitudes of an acute angled triangle ABC.

Prove that 1/AD + 1/BE > 1/CF.

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- Oct 17th 2010, 02:20 AMakshay1995related to triangles
Q. AD,BE,CF are the altitudes of an acute angled triangle ABC.

Prove that 1/AD + 1/BE > 1/CF. - Oct 17th 2010, 01:51 PMemakarov
Multiply this inequality by twice the area of ABC.

- Oct 18th 2010, 08:50 AMakshay1995
sorry , i couldn't understand. Can u please post the solved answer?

- Oct 18th 2010, 09:50 AMemakarov
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.