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In an acute triangle with angles $ A, B $ and $ C $, show that

$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $

I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small. Also, as $ A \to \frac{\pi}{2}, A+B+C \to \pi $ (as we want), but $ \cos {A} \cos {B} \cos {C} \to 0 $.

How can I proceed?

Dela Corte
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4 Answers4

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$$y=2\cos A\cos B\cos C=[\cos(A-B)+\cos(A+B)]\cos C=[\cos(A-B)-\cos C]\cos C$$

$$\implies\cos^2C-\cos(A-B)\cos C+y=0$$ which is Quadratic Equation in $\cos C$

As $C$ is real $\implies\cos C$ is real,

the discriminant $\cos^2(A-B)-4y\ge0\iff y\le\dfrac{\cos^2(A-B)}4\le\dfrac14$

lab bhattacharjee
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  • "As $C\iff \cos C$ is real": what does this mean? – TonyK Sep 30 '14 at 19:08
  • @TonyK, Sorry for the confusion – lab bhattacharjee Sep 30 '14 at 19:10
  • Well, it's still false. If $C=i$, then $\cos C$ is real, even though $C$ isn't. – TonyK Sep 30 '14 at 19:19
  • @TonyK, Is is technically correct ? – lab bhattacharjee Sep 30 '14 at 19:22
  • I don't know if that last line of yours is technically correct. You use more of those arrow symbols than everybody else on MSE put together, and sometimes I get the feeling that you are misusing them $-$ "$\implies$" is a verb , for instance, not a conjunction. And do you really want to say "$(y \le \cos^2(A-B)/4\le 1/4)$ implies $(\cos^2(A-B)-4y\ge0)$"? I get the feeling that there's something wrong there, but I can't quite put my finger on it. – TonyK Sep 30 '14 at 19:29
  • @TonyK, But I think this is stronger and when applied properly, shows better understanding – lab bhattacharjee Sep 30 '14 at 19:30
  • "$\implies$" is not the same as "therefore". I know that in the United Kingdom in the 1970's, students a little younger than I were taught in school that they mean the same thing. But wiser heads prevailed, and this particular phase only lasted two or three years. – TonyK Sep 30 '14 at 19:34
  • But if you think of $\cos(C)$ as variable, then $y$ is not a constant. – Redsbefall Jan 13 '20 at 04:11
  • @Arief, yes,$y$ is definitely not a constant, that's why we are trying to maximize it. Even if $y$ is not a constant, it forms the discriminant of the quadratic equation which must obey the rules of discriminant – lab bhattacharjee Jan 13 '20 at 04:24
  • @Arief, it will be cleaner to write $$\cos^2(A-B)-4y=(2\cos C-\cos(A-B))^2$$ which must be $\ge0$ – lab bhattacharjee Jan 13 '20 at 04:27
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Homage a lab

$2\cos A\cos B\cos C=(\cos(A+B)+\cos(A-B))\cos C=(\cos(A-B)-\cos C)\cos C=\frac{1}{4}\cos^2(A-B)-(\cos C-\frac{1}{2}\cos(A-B))^2\le\frac{1}{4}$

almagest
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0

We need to prove that $$\prod_{cyc}\frac{a^2+b^2-c^2}{2ab}\leq\frac{1}{8}$$ or $$a^2b^2c^2\geq\prod_{cyc}(a^2+b^2-c^2)$$ or $$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)\geq0,$$ which is Schur.

Done!

Michael Rozenberg
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NTS $\cos A\cos B\cos C \leq 1/8$ Now, since $ABC$ is a triangle, only one of the angles can possibly be equal or greater than $\pi/2$. So, without loss of generality, let $B, C<\pi/2$. Then $\cos B>0,\cos C>0.$

Case 1: $A\ge\pi/2$ $\Rightarrow \cos A<0 \Rightarrow \cos A\cos B\cos C<0<1/8$
Case 2: $A<\pi/2$ $\Rightarrow \cos A>0$
Then, by the AM-GM inequality: $$\sqrt[3]{\cos A\cos B\cos C}\leq {\frac 1 3}(\cos A+\cos B+\cos C)$$ Now, consider the function $f(x)=\cos x$ where $x \in(0,\pi/2)$. $f$ is a concave function because $f''(x)=-\cos x>0$ for $x\in(0,\pi/2)$. Then, by the Jenson's inequality: $$f(\frac1 3A +\frac 1 3B+\frac 13 C)\ge\frac 13f(A) +\frac13f(B)+\frac13f(C) \\$$ Then:$$\cos(\frac \pi3)\ge \frac13(\cos A+\cos B+\cos C)$$ Which is: $$\frac12\ge\frac13(\cos A+\cos B+\cos C)$$ Combining that result with the result from the AM-GM inequality, we get: $$\sqrt[3]{\cos A\cos B\cos C}\leq {\frac 1 3}(\cos A+\cos B+\cos C)\le\frac12$$ Cube each portion:$$\cos A\cos B\cos C\leq {\frac 1 {27}}(\cos A+\cos B+\cos C)^3\le\frac18$$

$$\therefore \cos A\cos B\cos C\leq \frac18$$

M. Yates
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