In a triangle ABC, prove that 1<cosA+cosB+cosC< or equal to 3/2

[anemone]

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 only hint, it would be much appreciated.

Thanks.

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

Thanks anyway.

 

[alexmahone]

Hint: Prove that $\displaystyle \cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.

 

[anemone]

I'm going to prove it by contradiction to show that in any triangle ABC, $1<cosA+cosB+cosC$ is correct.

First, I let $cosA+cosB+cosC<1$.

$cosA+cosB<1-cosC$

$2cos\frac{A+B}{2}cos\frac{A-B}{2}<1-(1-2sin^2 \frac{C}{2})$

$2sin\frac{C}{2}cos\frac{A-B}{2}-2sin^2 \frac{C}{2}<0$

$(2sin\frac{C}{2})(cos\frac{A-B}{2}-sin\frac{C}{2})<0$

Since $sin\frac{C}{2}>0$ which also means $2sin\frac{C}{2}>0$, and to have $(2sin\frac{C}{2})(cos\frac{A-B}{2}-sin\frac{C}{2})<0$, the following must be true:
$(cos\frac{A-B}{2}-sin\frac{C}{2})<0$

Thus,

$(cos\frac{A-B}{2}-sin\frac{C}{2})<0$

$(cos\frac{A-B}{2}-cos\frac{A+B}{2})<0$

$sinA<0$

This is a false statement and showing our assumption ($cosA+cosB+cosC<1$) leads to contradiction, thus concluding
$cosA+cosB+cosC>1$ must be true.:)

How does this sound to you?

 

[alexmahone]

I'm going to prove it by contradiction to show that in any triangle ABC, $1<cosA+cosB+cosC$ is correct.

First, I let $cosA+cosB+cosC<1$.

You should let $\cos A+\cos B+\cos C\le 1$.

$(cos\frac{A-B}{2}-cos\frac{A+B}{2})\le 0$

$sinA\le 0$

The second line should read $\displaystyle 2\sin\frac{A}{2}\sin\frac{B}{2}\le 0$.

 

[anemone]

Still I can't get thing right!(Sadface)

You should let $\cos A+\cos B+\cos C\le 1$.

OK. All right.

The second line should read $\displaystyle 2\sin\frac{A}{2}\sin\frac{B}{2}\le 0$.

Ah!
I'm so tired of all this and I thought I saw $2sin\frac{A}{2}cos\frac{A}{2}$!(Angry)