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

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The discussion centers on proving the inequality \( \cos A + \cos B + \cos C \leq \frac{3}{2} \) given that \( A + B + C = \pi \). The user applies Jensen's Inequality, demonstrating that for acute angles \( A \) and \( B \), the relationship \( \cos \frac{A+B}{2} \geq \frac{\cos A + \cos B}{2} \) holds. The proof is completed by manipulating trigonometric identities and confirming that the maximum value of the sum of cosines is indeed \( \frac{3}{2} \). The user also confirms that considering \( A \) and \( B \) as acute angles while ignoring \( C \) is valid due to the properties of triangles.

PREREQUISITES
  • Understanding of Jensen's Inequality in the context of convex functions
  • Knowledge of trigonometric identities and properties of angles in a triangle
  • Familiarity with the cosine function and its behavior in the interval \( (0, \frac{\pi}{2}) \)
  • Ability to manipulate algebraic expressions involving trigonometric functions
NEXT STEPS
  • Study the application of Jensen's Inequality in various mathematical contexts
  • Explore the properties of convex functions and their implications in optimization problems
  • Learn about trigonometric identities and their proofs in triangle geometry
  • Investigate other inequalities involving trigonometric functions, such as the Cauchy-Schwarz inequality
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Mathematicians, students studying trigonometry, and anyone interested in inequalities and their proofs in geometry.

anemone
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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.
 
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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.
 

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