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

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In summary, it is possible to solve the given question by assuming A and B are acute angles and ignoring angle C from the start. By letting f(x)=cosx and observing the convex shape of the curve of f(x)=cos x in the interval $x\in (0,\frac{\pi}{2})$, Jensen's inequality can be applied without any issues. This leads to the conclusion that $ cosA+cosB+cosC\leq \frac{3}{2}$, with the proof being considered 'neat' by the person asking the question. The phrase 'obviously' is also commonly used in mathematical discussions, but it may not always be considered true or helpful.
  • #1
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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  • #2
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!
 
  • #3
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.
 

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

1. How does Jensen's Inequality relate to cosA+cosB+cosC ≤ 3/2?

Jensen's Inequality is a mathematical concept that states that for a convex function, the average of the function is less than or equal to the function of the average. In this case, the average of cosA, cosB, and cosC is equal to (cosA+cosB+cosC)/3, and using Jensen's Inequality, we can prove that this average is less than or equal to cos((A+B+C)/3), which is equal to cos(π/3) or 1/2. Therefore, cosA+cosB+cosC ≤ (cosA+cosB+cosC)/3 ≤ 1/2, and by multiplying both sides by 3, we get the inequality cosA+cosB+cosC ≤ 3/2.

2. What is the significance of proving cosA+cosB+cosC ≤ 3/2 with Jensen's Inequality?

Proving this inequality with Jensen's Inequality is significant because it provides an elegant and rigorous proof for a mathematical statement that may seem intuitive but is not immediately obvious. Additionally, the use of Jensen's Inequality allows us to generalize this result to any convex function, not just trigonometric functions.

3. Can you explain the concept of convexity in Jensen's Inequality?

A function is considered convex if for any two points on the function's graph, the line segment connecting those points lies entirely above the function. In other words, the function is "curving upwards" and does not have any "dips" or "valleys." This concept is important in Jensen's Inequality because it guarantees that the average of the function is less than or equal to the function of the average.

4. How is the convexity of cosA+cosB+cosC used in the proof of the inequality?

In the proof, we use the fact that the cosine function is convex on the interval [0,π]. This means that for any three angles A, B, and C, the average of cosA, cosB, and cosC is less than or equal to cos((A+B+C)/3). We then substitute (A+B+C)/3 with π/3, which is the maximum value of the cosine function on the interval [0,π], and we get the inequality cosA+cosB+cosC ≤ (cosA+cosB+cosC)/3 ≤ 1/2, which is equivalent to cosA+cosB+cosC ≤ 3/2.

5. Can Jensen's Inequality be used to prove other trigonometric inequalities?

Yes, Jensen's Inequality can be applied to other trigonometric inequalities involving the sum of trigonometric functions. For example, it can be used to prove the inequality sinA+sinB+sinC ≤ 3√3/2, which is known as the Erdős–Mordell inequality. It can also be used to prove other types of inequalities in different mathematical contexts, such as in probability theory and functional analysis.

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