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Trigonometry Using of Jensen's Inequality to prove cosA+cosB+cosC less than or equal to 3/2 (where A+B+C=pi)

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,712
Hi,
Given $ A+B+C=\pi$, I need to prove $ cosA+cosB+cosC\leq \frac{3}{2}$.

I wish to ask if my following reasoning is correct.
First, I think of the case where A and B are acute angles, then I can use the Jensen's Inequality to show that the following is true.
$ cos\frac{A+B}{2}\geq \frac{cosA+cosB}{2}$
Carrying on with the working, I get
$ sin\frac{C}{2}\geq \frac{cosA+cosB}{2}$
$ 2sin\frac{C}{2}\geq cosA+cosB$
$ cosA+cosB\leq 2sin\frac{C}{2}$
$ cosA+cosB+cosC\leq 2sin\frac{C}{2}+cosC$
$ cosA+cosB+cosC\leq 2sin\frac{C}{2}+1-2sin^2C$
Completing square the RHS to obtain
$ cosA+cosB+cosC\leq -2(sin\frac{C}{2}-\frac{1}{2})^2+\frac{3}{2}$

Now, it's obvious to see that $ cosA+cosB+cosC\leq \frac{3}{2}$

My question is, can I solve the question by thinking A and B are acute angles and ignore the angle C right from the start so that I can let f(x)=cosx and notice that the curve of f(x)=cos x in the interval $x\in (0,\frac{\pi}{2})$ take the convex shape which in turn I can apply the Jensen's inequality without a problem?

Thanks.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
My question is, can I solve the question by thinking A and B are acute angles and ignore the angle C right from the start so that I can let f(x)=cosx and notice that the curve of f(x)=cos x in the interval $x\in (0,\frac{\pi}{2})$ take the convex shape which in turn I can apply the Jensen's inequality without a problem?
Yes: a triangle can have at most one obtuse angle, and the other two (acute) angles must necessarily be adjacent. Without loss of generality (favourite math phrase), take them to be A and B.

And BTW that is a very neat proof!
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,712
Yes: a triangle can have at most one obtuse angle, and the other two (acute) angles must necessarily be adjacent. Without loss of generality (favourite math phrase), take them to be A and B.
Thanks, Opalg.
And I think 'obviously' is also one of the favourite math phrase too!
But unfortunately that is a phrase to which I considered not so true and annoying some (if not most) of the time.:rolleyes:

And BTW that is a very neat proof!
:)
Thanks. Seriously, your compliment just made my day.