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

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The discussion focuses on proving the inequality \(1 < \cos A + \cos B + \cos C \leq \frac{3}{2}\) for triangle ABC. Participants explore methods to demonstrate the upper bound using techniques like completing the square and sum-to-product formulas, while also referencing the sine function's properties. A significant breakthrough occurs when one contributor suggests a proof by contradiction to establish that \(\cos A + \cos B + \cos C > 1\). The conversation highlights the challenges faced in the proof process, particularly in manipulating trigonometric identities correctly. Ultimately, the contributors are engaged in refining their approaches to solidify their understanding of the inequality.
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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 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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Hint: Prove that $\displaystyle \cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.
 
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?
 
anemone said:
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 $\displaystyle 2\sin\frac{A}{2}\sin\frac{B}{2}\le 0$.
 
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Still I can't get thing right!(Sadface)

Alexmahone said:
You should let $\cos A+\cos B+\cos C\le 1$.
OK. All right.
Alexmahone said:
The second line should read $\displaystyle 2\sin\frac{A}{2}\sin\frac{B}{2}\le 0$.
Ah!:mad:
I'm so tired of all this and I thought I saw $2sin\frac{A}{2}cos\frac{A}{2}$!(Angry)
 
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