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# 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

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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.

Staff - 3 years, 5 months ago

"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.

- 3 years, 5 months ago

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.

- 3 years, 5 months ago

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

- 3 years, 5 months ago

use AP > GP..

- 3 years, 5 months ago

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.

- 3 years, 5 months ago

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

- 3 years, 5 months ago