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
 A: One of my (who happens to be the problem's author) solutions:
We'll prove the following equivalent statement:
If
$$x+y+z=\frac\pi4;~~~-\frac\pi4\leq x,y,z\leq\frac\pi4,$$
then
$$-1\leq2\sqrt2\sin x\sin y\sin z\leq\frac{\left(\sqrt3-1\right)^3}8.$$
Study the equality cases.
WLOG assume that $x\leq y\leq z.$ If $xyz=0,$ then we have nothing to prove.
Otherwise, if $y<0$ then $z>\frac\pi4,$ which is false. So $y>0.$
Let's prove the right inequality. If $x<0$ then we are done. So let $x>0.$
In this case the right inequality is proved, due to Jensen's inequality. Equality iff $x=y=z,$ or $ABC$ is equilateral.
We prove now the left one. If $x>0,$ then we are done. So let $x<0.$
We need to prove
$$2\sqrt2\sin (-x)\sin y\sin z\leq1.$$
But this is obviously true, since $0<\sin (-x),\sin y,\sin z\leq\frac1{\sqrt2}.$
The equality can be achieved for $x=-\frac\pi4;y=z=\frac\pi4$ for example. The proof is complete.
References: https://artofproblemsolving.com/community/c6h2815170p24872115
The general problem is:If $\Delta ABC$ is acute angled and $\alpha\in\left(0,\frac\pi3\right]$ is fixed,
then
$$\max\left\{\sin^2\alpha\cos\alpha;\cos^3\left(\frac\pi3+\alpha\right)\right\}\geq\prod_{\text{cyc}}{\cos\left(A+\alpha\right)}\geq\min\left\{-\sin\alpha\cos^2\left(\frac\pi4+\alpha\right);\cos^3\left(\frac\pi3+\alpha\right)\right\}.$$
(C. Mateescu and L. Giugiuc, 2022)
References: https://artofproblemsolving.com/community/c6t243f6h2826343_cosines_product_with_parameter
A: $\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right) = -2\sqrt{2} \prod_{cyc} (A - \pi / 4)$.
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.
A: Nice. This is one of my recent creations. I'll soon publish an article regarding the general config, together with a friend
