Asking About Best Upper Bound And lowerBound Find the best lower and upper bounds for
$$\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right),$$
$1)~~$ overall acute-angled $\Delta ABC.$
first of all we know that $\cos(A)-\sin(A) = -\sqrt{2}\sin(A - \frac{\pi}{4})$
$\Pi { -\sqrt{2}\sin(A - \frac{\pi}{4})}=-2\sqrt{2}*\Pi{(\sin(A-\frac{\pi}{4}))}$
$-\frac{1}{2}*\sqrt{2}<\sin(A-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
$-\frac{1}{2}*\sqrt{2}<\sin(B-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
$-\frac{1}{2}*\sqrt{2}<\sin(C-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
Then Lagrange? or there is any other method you all could suggest
Edit:
For Everyone who wanna post or finish using every possible method just write it down here

$\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right) = -2\sqrt{2} \prod_{cyc} (A - \pi / 4)$.
WLOG, $\pi / 2 > A \geq B \geq C > 0$.
Easy to see $\pi /3 \leq A < \pi/2$, $\pi/4 < B \leq A$, and $0 < C \leq B$.
Clearly, upper bound is achieved when $\sin(C - \pi/4) < 0$, i.e., $0 < C < \pi/4$.
Fix $C < \pi/4$, we have $\pi/2 - C/2 \leq A < \pi /2$ and $B = \pi - A -C$.
When $C$ is fixed, the maximum is achieved when $\sin(A - \pi/4)\sin(B-\pi/4)$ is maximzed, where
$$\sin(A - \pi/4)\sin(B-\pi/4) = \sin(A - \pi/4)\sin(3\pi/4 - A - C).$$
Easy to see the above equation is maximized when $A - \pi/4 = 3\pi/4 - A - C = \pi/2 - C/2$, i.e., $A = 3\pi/4 - C/2$ and the maximum value is $\sin^2(\pi/2 - C/2)$.
The next step is to maximize
$$-2\sqrt{2} \sin^2(\pi/2 - C/2)\sin(C-\pi/4),$$
where $0 < C < \pi/4$.
Regarding the lower bound, we shall have $\pi/3 \geq C > \pi/4$ and $\sin(C-\pi/4) > 0$.
This time again we need to maximize $\sin(A - \pi/4)\sin(B-\pi/4)$ in a similar reasoning.
