$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ 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?

$$y=2\cos A\cos B\cos C=[\cos(A-B)+\cos(A+B)]\cos C=[\cos(A-B)-\cos C]\cos C$$
$$\implies\cos^2C-\cos(A-B)\cos C+y=0$$ which is Quadratic Equation in $\cos C$
As $C$ is real $\implies\cos C$ is real, 
the discriminant $\cos^2(A-B)-4y\ge0\iff y\le\dfrac{\cos^2(A-B)}4\le\dfrac14$
