Solve Very Hard Inequality: a,t in (0,1)

[Nedeljko]

Hi, folks,

I need the proof of following inequality for a,t\in(0,1):

Weaker form: \ln(1+t^a)\geq a\ln^a(1+t).

Stronger form: \ln(1+t^a)\geq\frac{1+a}2\ln^a(1+t).

 

[JG89]

For the weaker form, on the right hand side do you mean (1 + t) ^ a?

 

[vishal_garg]

Can you please rite your ques. in a proper way...

 

[HallsofIvy]

You mean using words like "please" and "rite"? Other than that, what is wrong with the way he wrote it?

 

[vishal_garg]

first of all it's not an english forum site..
n secondly i ain't sayin that to you...
you better behave well..

 

[sutupidmath]

first of all it's not an english forum site..
n secondly i ain't sayin that to you...
you better behave well..

Yes, it is not an english forum site, but you have to know what you are saying, and actually what you need to do is have respect for the members here, especially for our honored experts like HallsofIvy. SO, if anyone here needs to well behave, than that is most certainly YOU!

Regards!

 

[vishal_garg]

i didn't understood his question so i wrote that blog.
i wasn't abusive..
i just requested him to rewrite his question so i can understand it...
i was behaving well but your honored expert was having a problem with that thing..
i mean i am only 16 years old and yet i am trying to solve that problem and you people are not supporting me instead using harsh language with me..

 

[Mentallic]

you better behave well..

This is what set them off, anyone with a right mind would take it as a threat/warning, which is clearly inappropriate.

Sorry I cannot help, but find this question very interesting: I will now go and search for meaning to "weaker" and "stronger" forms of inequalities which you have used.

 

[Nedeljko]

I proved the stronger form

\ln(1+t^a)>\ln^{1-a}2\ln^a(1+t)

or equivalently

\log_2(1+t^a)>\log_2^a(1+t)

and some generalisations.