Nice Question:
let x∈[0,2π], show that:
sinsinsinsinx≤45coscoscoscosx?
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 BBSI post this problem solution
case1: if x∈[π,2π],then
coscoscoscosx>0,sinsinsinsinx≤0
so
coscoscoscosx>sinsinsinsinxcase2: if x∈[0,π2],then we have
cosx+sinx≤√2<π2⟹0≤cosx<π2−sinx
so
coscosx>cos(π2−sinx)=sinsinx
sincosx<sin(π2−sinx)=cossinx
then
coscoscosx<cossinsinx
so
coscoscosx+sinsinsinx<cossinsinx+sinsinsinx<π2
so
coscoscosx<π2−sinsinsinx
then
coscoscoscosx>cos(π2−sinsinsinx)=sinsinsinsinx
case3: if x∈(π2,π),then lety=x−π2,so
coscoscossiny>sinsincossiny
and since
f(t)=sinsint is increasing,then
f(cossiny)>f(sincosy)⟹sinsincossiny>sinsinsincosy
so
coscoscossiny>sinsinsincosy
so
coscoscoscosx>sinsinsinsinxBut I found this 45 maybe is strong,
so if x∈[π,2π],then we have
45coscoscoscosx≥0>sinsinsinsinxBut for the case x∈[0,π], I can't prove this
4coscoscoscosx≥5sinsinsinsinxThank 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 −1≤cosx≤1,that is cos1≤coscosx≤1,that is cos1≤coscoscosx≤coscos1.
Finally, we have,
0.6542≃coscoscos1≤coscoscoscosx≤coscos1.
Similarly, we have
−sinsinsinsin1≤sinsinsinsinx≤sinsinsin1≃0.6784...
If the equation has the solution, that is
sinx≥sin−1sin−1sin−1coscoscos1≃0.6086...
Thus, we have
|cosx|≤0.7835...
Therefore, coscos≥0.7013→coscoscosx≤0.7639→coscoscoscoscosx≥0.7221...
Thus, it is not possible to have sinsinsinsinsin≤0.6784. 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