1) Có lẽ là $\widehat{A}+\widehat{B}+\widehat{C}=\pi$:
Đặt $(\omega_1,\omega_2,\omega_3) = \left(\frac{A}{4},\frac{B}{4}, \frac{C}{4}\right)$ thì $\omega_1+\omega_2+\omega_3=\frac{\pi}{4}$.
Ta có: $\tan(\omega_1 + \omega_2) = \tan\left(\frac{\pi}{4} - \omega_3\right)$
$\frac{\tan \omega_1 + \tan\omega_2}{1-\tan \omega_1\tan\omega_2} = \frac{1 - \tan\omega_3}{1+\tan\omega_3}$
$\Rightarrow \sum_{i<j}\tan\omega_i\omega_j + \sum \tan\omega_i = 1 + \tan\omega_1\tan\omega_2\tan\omega_3$
$\Rightarrow (1+\tan\omega_1)(1+\tan\omega_2)(1+\tan\omega_3) = 2(1+\tan\omega_1\tan\omega_2\tan\omega_3)$.
2) Đặt $(x,y,z) = (\tan \omega_1, \tan\omega_2, \tan\omega_3)$.
Ta có $xyz + 1 = xy + yz +zx + x+y+z\geq 3\sqrt[3]{(xyz)^2} + 3\sqrt[3]{xyz}$
$\Rightarrow \sqrt[3]{xyz}\leq 2-\sqrt{3}$ (Do $\sqrt[3]{xyz} < 1$)
$\Rightarrow P = 2\left(1 + xyz\right) \leq 6(5-\sqrt{3})$.
Vậy $\max P = 6(5-\sqrt{3})$, xảy ra khi $A=B=C=\frac{\pi}{6}$.