How prove this inequality sinsinsinsinx≤45coscoscoscosx\sin{\sin{\sin{\sin{x}}}}\le\frac{4}{5}\cos{\cos{\cos{\cos{x}}}}

Question 1

Nice Question:

let $x\in [0,2\pi]$ , show that:

$\sin{\sin{\sin{\sin{x}}}}\le\dfrac{4}{5}\cos{\cos{\cos{\cos{x}}}}?$

I know this follow famous problem(1995 Russia Mathematical olympiad)

$\sin{\sin{\sin{\sin{x}}}}<\cos{\cos{\cos{\cos{x}}}}$

This problem solution can see :http://iask.games.sina.com.cn/b/19776980.html
and everywhere have solution in china BBS

I post this problem solution

case1: if $x\in[\pi,2\pi]$ ,then
$\cos{\cos{\cos{\cos{x}}}}>0,\sin{\sin{\sin{\sin{x}}}}\le 0$
so
$\cos{\cos{\cos{\cos{x}}}}>\sin{\sin{\sin{\sin{x}}}}$

case2: if $x\in[0,\dfrac{\pi}{2}]$ ,then we have
$\cos{x}+\sin{x}\le\sqrt{2}<\dfrac{\pi}{2}\Longrightarrow 0\le \cos{x}<\dfrac{\pi}{2}-\sin{x}$
so
$\cos{\cos{x}}>\cos{\left(\dfrac{\pi}{2}-\sin{x}\right)}=\sin{\sin{x}}$
$\sin{\cos{x}}<\sin{\left(\dfrac{\pi}{2}-\sin{x}\right)}=\cos{\sin{x}}$
then
$\cos{\cos{\cos{x}}}<\cos{\sin{\sin{x}}}$
so
$\cos{\cos{\cos{x}}}+\sin{\sin{\sin{x}}}<\cos{\sin{\sin{x}}}+\sin{\sin{\sin{x}}}<\dfrac{\pi}{2}$
so
$\cos{\cos{\cos{x}}}<\dfrac{\pi}{2}-\sin{\sin{\sin{x}}}$
then
$\cos{\cos{\cos{\cos{x}}}}>\cos{\left(\dfrac{\pi}{2}-\sin{\sin{\sin{x}}}\right)}=\sin{\sin{\sin{\sin{x}}}}$
case3: if $x\in (\dfrac{\pi}{2},\pi)$ ,then let

$y=x-\dfrac{\pi}{2}$ ,so
$\cos{\cos{\cos{\sin{y}}}}>\sin{\sin{\cos{\sin{y}}}}$
and since
$f(t)=\sin{\sin{t}}$ is increasing,then
$f(\cos{\sin{y}})>f(\sin{\cos{y}})\Longrightarrow \sin{\sin{\cos{\sin{y}}}}>\sin{\sin{\sin{\cos{y}}}}$
so
$\cos{\cos{\cos{\sin{y}}}}>\sin{\sin{\sin{\cos{y}}}}$
so
$\cos{\cos{\cos{\cos{x}}}}>\sin{\sin{\sin{\sin{x}}}}$

But I found this $\dfrac{4}{5}$ maybe is strong,

so if $x\in[\pi,2\pi]$ ,then we have
$\dfrac{4}{5}\cos{\cos{\cos{\cos{x}}}}\ge 0>\sin{\sin{\sin{\sin{x}}}}$

But for the case $x\in [0,\pi]$ , I can't prove this
$4\cos{\cos{\cos{\cos{x}}}}\ge 5\sin{\sin{\sin{\sin{x}}}}$

Thank you very much！

Question 2

We still start from the original Russian Olympiad Problem:
$\cos \cos \cos \cos x> \sin \sin \sin \sin x$ . It could have another numerical proof simply by doing in a calculator:

We have $-1\leq \cos x \leq 1, \text{that is }\cos 1\leq \cos \cos x\leq 1, \text{that is }\cos 1 \leq \cos \cos \cos x \leq \cos \cos 1$ .

Finally, we have,

$~0.6542 \simeq \cos \cos \cos 1\leq \cos \cos \cos \cos x \leq \cos \cos 1.$

Similarly, we have

$-\sin \sin \sin \sin 1 \leq \sin \sin \sin \sin x \leq \sin \sin \sin 1 \simeq0.6784...$

If the equation has the solution, that is

$\sin x\geq \sin^{-1} \sin^{-1}\sin ^{-1} \cos \cos \cos 1 \simeq 0.6086...$

Thus, we have
$|\cos x|\leq 0.7835...$

Therefore, $\cos \cos \geq 0.7013 \to \cos \cos \cos x \leq 0.7639 \to \cos \cos \cos \cos \cos x \geq 0.7221...$

Thus, it is not possible to have $\sin \sin \sin \sin \sin \leq 0.6784.$ The inequality holds.

So, the $\frac{4}{5}$ is still not a strong constant, and inequality proof as similar.

How prove this inequality sinsinsinsinx≤45coscoscoscosx\sin{\sin{\sin{\sin{x}}}}\le\frac{4}{5}\cos{\cos{\cos{\cos{x}}}}

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