Classic Trig Inequality

Prove that for all triangles with angles α,β,γ\alpha,\beta,\gamma that cosαcosβcosγ18\cos\alpha\cos\beta\cos\gamma\le \dfrac{1}{8}

I just want to see as many solutions as possible.

#Geometry #Trigonometry #Triangle #Proofs #Inequality

Note by Daniel Liu
6 years, 12 months ago

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Comments

Show that lncosx \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 - 6 years, 12 months ago

"As many solutions as possible"? Okay, here's another approach. Let xx be the incremental variable, and we find the series expansion of Cos(α+x)Cos(β+x)Cos(γ2x)Cos(\alpha +x)Cos(\beta +x)Cos(\gamma -2x), which works out to, after simplification

Cos(α)Cos(β)Cos(γ)+(12Sin(2α)+12Sin(2β)Sin(2γ))x+Ox2+...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 OO indicates higher order coefficients. For the expression to be an extrema, the 2nd2nd term must vanish, that is

12Sin(2α)+12Sin(2β)Sin(2γ))=0\frac { 1 }{ 2 } Sin(2\alpha )+\frac { 1 }{ 2 } Sin(2\beta )-Sin(2\gamma ))=0

But for symmetry reasons, this must vanish also

12Sin(2α)Sin(2β)+12Sin(2γ))=0\frac { 1 }{ 2 } Sin(2\alpha )-Sin(2\beta )+\frac { 1 }{ 2 } Sin(2\gamma ))=0, or

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

Adding this to the first, we quickly find that

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

and that eventually α=β=γ=π3\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 α+β+γ=π\alpha +\beta +\gamma =\pi

18Cos(α)Cos(β)Cos(γ)1-8Cos(\alpha )Cos(\beta )Cos(\gamma )

(Sin(αβ))2+(Cos(αβ)2Cos(γ))2{ (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 00. Hence, the inequality follows.

Michael Mendrin - 6 years, 11 months ago

use AP > GP..

Pradeep Ch - 6 years, 12 months ago

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Thanks!

Solution:

We see that cos(α+β+γ3)cosα+cosβ+cosγ3cosαcosβcosγ3\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 α+β+γ=π\alpha+\beta+\gamma=\pi socosαcosβcosγ3cosπ3=12\sqrt[3]{\cos\alpha\cos\beta\cos\gamma}\le \cos \dfrac{\pi}{3}=\dfrac{1}{2}

Thus cosαcosβcosγ18\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, cosαcosβcosγ<0\cos\alpha\cos\beta\cos\gamma < 0 so the inequality still holds.

Daniel Liu - 6 years, 12 months ago

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

Pradeep Ch - 6 years, 12 months ago

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

Ramesh Goenka - 6 years, 11 months ago

Would this work?:

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

Thus we should have at least some two of α,β,γ \alpha, \beta, \gamma equal say α=γ \alpha=\gamma to maximize the LHS of the inequality. This also means β=1802α \beta=180-2\alpha

Now we wish to maximize cos2αcos(1802α)=cos2αcos2α \cos^2{\alpha}\cos{(180-2\alpha)}=-\cos^2{\alpha}\cos{2\alpha} .

We rewrite as 1+cos2α2cos2α -\frac{1+\cos{2\alpha}}{2} \cdot \cos{2\alpha} . As a quadratic in cos2α\cos{2\alpha} the maximum is 1/8 when cos2α=1/2 \cos{2\alpha}=-1/2 . This is equivalent to an equilateral triangle.

Kaan Dokmeci - 6 years, 11 months ago
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