Waste less time on Facebook — follow Brilliant.
×

Classic Trig Inequality

Prove that for all triangles with angles \(\alpha,\beta,\gamma\) that \[\cos\alpha\cos\beta\cos\gamma\le \dfrac{1}{8}\]

I just want to see as many solutions as possible.

Note by Daniel Liu
3 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Show that \( \ln \cos x \) is a concave function and apply Jensens.

This is, in essence, combining the 2 steps of your solution into 1.

Calvin Lin Staff - 3 years, 5 months ago

Log in to reply

"As many solutions as possible"? Okay, here's another approach. Let \(x\) be the incremental variable, and we find the series expansion of \(Cos(\alpha +x)Cos(\beta +x)Cos(\gamma -2x)\), which works out to, after simplification

\(Cos(\alpha )Cos(\beta )Cos(\gamma )+(\frac { 1 }{ 2 } Sin(2\alpha )+\frac { 1 }{ 2 } Sin(2\beta )-Sin(2\gamma ))x+O{ x }^{ 2 }+...\)

where \(O\) indicates higher order coefficients. For the expression to be an extrema, the \(2nd\) term must vanish, that is

\(\frac { 1 }{ 2 } Sin(2\alpha )+\frac { 1 }{ 2 } Sin(2\beta )-Sin(2\gamma ))=0\)

But for symmetry reasons, this must vanish also

\(\frac { 1 }{ 2 } Sin(2\alpha )-Sin(2\beta )+\frac { 1 }{ 2 } Sin(2\gamma ))=0\), or

\(Sin(2\alpha )-2Sin(2\beta )+Sin(2\gamma ))=0\)

Adding this to the first, we quickly find that

\(Sin(2\alpha )=Sin(2\beta )\)

and that eventually \(\alpha =\beta =\gamma =\dfrac { \pi }{ 3 } \), and the rest more or less follows. At this point, we've proven that this is the only extrema possible (with some caveats), but we haven't proven that it's a maxima. This is just a different approach, which can be used in other inequality problems like this. This is all part of "finding where the extrema are hiding?" problem, if you can't quite plot the landscape. If you can't see them, then feel them.

ADDED: In the spirit of "coming up with more proofs", here's a short proof. The following two expressions are equivalent, provided \(\alpha +\beta +\gamma =\pi \)

\(1-8Cos(\alpha )Cos(\beta )Cos(\gamma )\)

\({ (Sin(\alpha -\beta )) }^{ 2 }+{ (Cos(\alpha -\beta )-2Cos(\gamma )) }^{ 2 }\)

But notice that the 2nd expression, being a sum of squares, has to be positive or \(0\). Hence, the inequality follows.

Michael Mendrin - 3 years, 5 months ago

Log in to reply

Would this work?:

Let \( f(x)=\cos{x}\cos{(K-x)} \) where \( 0<K<180 \) is constant. We can rewrite f as \( f(x)=1/2\cos{(x+K-x)}+1/2\cos{(x-(K-x))}=1/2\cos{K}+1/2\cos{(2x-K)} \) which is maximized when \( x=K/2 \) (x is less than K for our intentions.)

Thus we should have at least some two of \( \alpha, \beta, \gamma \) equal say \( \alpha=\gamma \) to maximize the LHS of the inequality. This also means \( \beta=180-2\alpha \)

Now we wish to maximize \( \cos^2{\alpha}\cos{(180-2\alpha)}=-\cos^2{\alpha}\cos{2\alpha} \).

We rewrite as \( -\frac{1+\cos{2\alpha}}{2} \cdot \cos{2\alpha} \). As a quadratic in \(\cos{2\alpha} \) the maximum is 1/8 when \( \cos{2\alpha}=-1/2 \). This is equivalent to an equilateral triangle.

Kaan Dokmeci - 3 years, 5 months ago

Log in to reply

take the measure of the angles as 60 degrees . :P

Ramesh Goenka - 3 years, 5 months ago

Log in to reply

use AP > GP..

Pradeep Ch - 3 years, 5 months ago

Log in to reply

Thanks!

Solution:

We see that \[\cos \left(\dfrac{\alpha+\beta+\gamma}{3}\right)\ge \dfrac{\cos\alpha+\cos\beta+\cos\gamma}{3}\ge \sqrt[3]{\cos\alpha\cos\beta\cos\gamma}\]

where the first inequality is Jensen's and the second is AM-GM.

But \(\alpha+\beta+\gamma=\pi\) so\[\sqrt[3]{\cos\alpha\cos\beta\cos\gamma}\le \cos \dfrac{\pi}{3}=\dfrac{1}{2}\]

Thus \[\cos\alpha\cos\beta\cos\gamma \le\dfrac{1}{8}\] as desired.

This solution works as long as the triangle is acute. However, when the triangle is obtuse, \(\cos\alpha\cos\beta\cos\gamma < 0\) so the inequality still holds.

Daniel Liu - 3 years, 5 months ago

Log in to reply

yeah.. this is exactly what i was talking about..

Pradeep Ch - 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...