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- Feb 14, 2012

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In a triangle ABC, prove that $1<cosA+cosB+cosC \leq \frac{3}{2} $.

One can easily prove that $cosA+cosB+cosC \leq 3/2 $, i.e. it can be proven to be true by

1. Using only the method of completing the square with no involvement of any inequality formula like Jensen's, AM-GM, etc.

2. By sum-to-product formulas and the fact that $sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} \leq \frac{1}{8} $

But no matter how hard I try, I couldn't prove the sum of the angles A, B and C to be greater than 1. In fact, I find myself always end up with the 'less than' sign.

If you can offer any help by giving

Thanks.

(Edit: I think I know how to prove it now: By contradiction!)

Thanks anyway.

One can easily prove that $cosA+cosB+cosC \leq 3/2 $, i.e. it can be proven to be true by

1. Using only the method of completing the square with no involvement of any inequality formula like Jensen's, AM-GM, etc.

2. By sum-to-product formulas and the fact that $sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} \leq \frac{1}{8} $

But no matter how hard I try, I couldn't prove the sum of the angles A, B and C to be greater than 1. In fact, I find myself always end up with the 'less than' sign.

If you can offer any help by giving

**only**hint, it would be much appreciated.Thanks.

(Edit: I think I know how to prove it now: By contradiction!)

Thanks anyway.

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