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

Nice Question:

let x[0,2π], show that:

sinsinsinsinx45coscoscoscosx?

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

sinsinsinsinx<coscoscoscosx

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[π,2π],then
coscoscoscosx>0,sinsinsinsinx0
so
coscoscoscosx>sinsinsinsinx

case2: if x[0,π2],then we have
cosx+sinx2<π20cosx<π2sinx
so
coscosx>cos(π2sinx)=sinsinx
sincosx<sin(π2sinx)=cossinx
then
coscoscosx<cossinsinx
so
coscoscosx+sinsinsinx<cossinsinx+sinsinsinx<π2
so
coscoscosx<π2sinsinsinx
then
coscoscoscosx>cos(π2sinsinsinx)=sinsinsinsinx
case3: if x(π2,π),then let

y=xπ2,so
coscoscossiny>sinsincossiny
and since
f(t)=sinsint is increasing,then
f(cossiny)>f(sincosy)sinsincossiny>sinsinsincosy
so
coscoscossiny>sinsinsincosy
so
coscoscoscosx>sinsinsinsinx

But I found this 45 maybe is strong,

so if x[π,2π],then we have
45coscoscoscosx0>sinsinsinsinx

But for the case x[0,π], I can't prove this
4coscoscoscosx5sinsinsinsinx

Thank you very much!

Answer

We still start from the original Russian Olympiad Problem:
coscoscoscosx>sinsinsinsinx. It could have another numerical proof simply by doing in a calculator:

We have 1cosx1,that is cos1coscosx1,that is cos1coscoscosxcoscos1.

Finally, we have,

 0.6542coscoscos1coscoscoscosxcoscos1.

Similarly, we have

sinsinsinsin1sinsinsinsinxsinsinsin10.6784...

If the equation has the solution, that is

sinxsin1sin1sin1coscoscos10.6086...

Thus, we have
|cosx|0.7835...

Therefore, coscos0.7013coscoscosx0.7639coscoscoscoscosx0.7221...

Thus, it is not possible to have sinsinsinsinsin0.6784. The inequality holds.

So, the 45 is still not a strong constant, and inequality proof as similar.

Attribution
Source : Link , Question Author : math110 , Answer Author : LorenMt

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