MHB Proving cosA+cosB+cosC ≤ 3/2 with Jensen's Inequality

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Inequality
AI Thread Summary
The discussion focuses on proving that cosA + cosB + cosC ≤ 3/2 given A + B + C = π, utilizing Jensen's Inequality. The reasoning begins by considering A and B as acute angles, allowing the application of Jensen's Inequality to derive relationships between the angles. The conclusion is reached by manipulating trigonometric identities and completing the square, ultimately demonstrating the inequality holds true. Participants agree that it is valid to assume A and B are acute while ignoring C, as a triangle can have at most one obtuse angle. The proof is acknowledged as neat and effective, with a light-hearted exchange about mathematical expressions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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.
 
Mathematics news on Phys.org
anemone said:
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!
 
Opalg said:
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:

Opalg said:
And BTW that is a very neat proof!
:)
Thanks. Seriously, your compliment just made my day.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top