Yff's Conjecture

Theorem
Let $\triangle ABC$ be a triangle.

Let $\omega$ be the Brocard angle of $\triangle ABC$.

Then:
 * $8 \omega^3 < ABC$

where $A, B, C$ are measured in radians.

Proof
The Abi-Khuzam Inequality states that


 * $\sin A \cdot \sin B \cdot \sin C \le \paren {\dfrac {3 \sqrt 3} {2 \pi} }^3 A \cdot B \cdot C$

The maximum value of $A B C - 8 \omega^3$ occurs when two of the angles are equal.

So taking $A = B$, and using $A + B + C = \pi$, the maximum occurs at the maximum of:


 * $\map f A = A^2 \paren {\pi - 2 A} - 8 \paren {\map \arccot {2 \cot A - \cot 2 A} }^3$

which occurs when:


 * $2 A \paren {\pi - 3 A} - \dfrac {48 \paren {\map \arccot {\frac 1 2 \paren {3 \cot A + \tan A} } }^2 \paren {1 + 2 \cos 2 A} } {5 + 4 \cos 2 A} = 0$

Also known as
Can also be seen referred to as the Yff conjecture.