Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.

If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.

I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.

On this topic we learned also about Cauchy inequality, but I have no experience with it.

The answer according to Mathematica is when $A=B=C=60$.

Any ideas?

Solution 1:

We may assume $\alpha\leq\beta\leq\gamma$. The product $p:=\cos\alpha\cos\beta\cos\gamma$ is negative iff $\gamma>{\pi\over2}$, so that the minimum value $p_{\min}=-1$ is attained when $\alpha=\beta=0$, $\gamma=\pi$, i.e., for a degenerate triangle.

In the search of $p_{\max}$ we may assume $\gamma\leq{\pi\over2}$, hence $0\leq\cos\gamma\leq1$. From $$2\cos\alpha\cos\beta=\cos(\alpha-\beta)+\cos(\alpha+\beta)\leq 1+\cos(\alpha+\beta)$$ it follows that $$p\leq{1\over2}(1-\cos\gamma)\cos\gamma={1\over2}\left({1\over4}-\left(\cos\gamma-{1\over2}\right)^2\right)\leq{1\over8}\ .$$ This proves $p_{\max}={1\over8}$, and this value is attained when $\gamma={\pi\over3}$ and $\alpha=\beta$, i.e., for an equilateral tringle.

Solution 2:

If $y=\cos A\cos B\cos C,$

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

As $A+B=\pi-C,\cos(A+B)=-\cos C$

On rearrangement we have $$\cos^2C-\cos C\cos(A-B)+2y=0$$

As $C$ is real, so will be $\cos C$

$\implies$ the discriminant $$\cos^2(A-B)-8y\ge0\iff y\le\dfrac{\cos^2(A-B)}8\le\dfrac18$$

The equality occurs if $\cos^2(A-B)=1\iff\sin^2(A-B)=0$

$\implies A-B=n\pi$ where $n$ is any integer

As $0<A,B<\pi, n=0\iff A=B$ and consequently $$\cos^2C-\cos C+2\cdot\dfrac18=0\implies \cos C=\dfrac12\implies C=\dfrac\pi3$$

$\implies A=B=\dfrac{A+B}2=\dfrac{\pi-C}2=\dfrac\pi3=C$