Asking About Best Upper Bound And lowerBound

Question

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

Answer 1

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

Answer 2

$\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right) = -2\sqrt{2} \prod_{cyc} (A - \pi / 4)$.

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.

Answer 3

Nice. This is one of my recent creations. I'll soon publish an article regarding the general config, together with a friend