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

In summary, the conversation discusses the proof of the inequality \ln(1+t^a)\geq a\ln^a(1+t) for a,t\in(0,1). The weaker form states that the inequality holds for a\ln^a(1+t) while the stronger form states that it holds for \frac{1+a}2\ln^a(1+t). The conversation also includes a request for clarification and a disagreement over appropriate behavior. One member offers to search for related inequalities and another claims to have proved the stronger form.
  • #1
Nedeljko
40
0
Hi, folks,

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

Weaker form: [tex]\ln(1+t^a)\geq a\ln^a(1+t)[/tex].

Stronger form: [tex]\ln(1+t^a)\geq\frac{1+a}2\ln^a(1+t)[/tex].
 
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  • #2
For the weaker form, on the right hand side do you mean (1 + t) ^ a?
 
  • #3
Can you please rite your ques. in a proper way...
 
  • #4
You mean using words like "please" and "rite"? Other than that, what is wrong with the way he wrote it?
 
  • #5
first of all it's not an english forum site..
n secondly i ain't sayin that to you...
you better behave well..
 
  • #6
vishal_garg said:
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!
 
  • #7
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..
 
  • #8
vishal_garg said:
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.
 
  • #9
I proved the stronger form

[tex]\ln(1+t^a)>\ln^{1-a}2\ln^a(1+t)[/tex]

or equivalently

[tex]\log_2(1+t^a)>\log_2^a(1+t)[/tex]

and some generalisations.
 
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FAQ: Solve Very Hard Inequality: a,t in (0,1)

What is an inequality?

An inequality is a mathematical statement that compares two quantities and indicates whether one is greater than, less than, or equal to the other.

What does it mean for a variable to be in the range of (0,1)?

A variable being in the range of (0,1) means that its value can only be between 0 and 1, but not including 0 and 1 themselves. This is also known as an open interval.

How do you solve a very hard inequality?

To solve a very hard inequality, you need to first simplify the expression as much as possible. Then, use algebraic manipulation and properties of inequalities to isolate the variable on one side of the inequality symbol. Finally, use test values and the properties of the inequality to determine the solution set.

Is the solution to an inequality always a single number?

No, the solution to an inequality can be a range of values. This is because inequalities represent a set of numbers that satisfy the given conditions, rather than a single specific value.

Why is it important to specify the range of a variable in an inequality?

Specifying the range of a variable in an inequality is important because it limits the possible values that the variable can take, making it easier to solve the inequality. It also ensures that the solution set is accurate and does not include any extraneous values.

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