Inequality proof problem of the day # 1

If x,yx,y are real numbers such that x2014+y2014=1x^{2014}+y^{2014}=1 then prove that:

(n=11007x2n+1x4n+1)(n=11007y2n+1y4n+1)<1(1x)(1y)\left(\sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left(\sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right)< \dfrac{1}{(1-x)(1-y)}

Note by Shivam Jadhav
3 years, 9 months ago

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There is a typo in the question. I will explain you what the typo is.

For suppose, take x=1x=-1 and y=0y=0. Then we get

n=11007x2n+1x4n+1=1007n=11007y2n+1y4n+1=10071(1x)(1y)=12 \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} = 1007 \\ \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} = 1007 \\ \dfrac{1}{\left(1-x \right) \left(1-y \right)} = \dfrac{1}{2}

So, according to our question it forces that 1007×1007<1/21007 \times 1007 < 1/2, which is impossible.


The question should be actually like this:

If xx, yyare real numbers such that x2014+y2014=1x^{2014}+y^{2014}=1 then prove that:

(n=11007x2n+1x4n+1)(n=11007y2n+1y4n+1)<1(1x)(1y)\left( \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right) < \dfrac{1}{\left(1-|x| \right) \left( 1-|y| \right)}


PROOF:

Let x=a|x| = a and y=b|y| = b. So, the given constraint changes into a2014+b2014=1a^{2014} + b^{2014} = 1.

Since, a4n+1>(a2n+1)22=12×(a2n+1)×(a2n+1)>an(a2n+1) a^{4n} + 1 > \dfrac{ \left( a^{2n} + 1 \right)^2}{2} = \dfrac{1}{2} \times \left( a^{2n} + 1 \right) \times \left( a^{2n} + 1 \right) > a^{n} \left( a^{2n} + 1 \right). Which implies that

1+a2n1+a4n<1ann=110071+a2n1+a4n<n=110071an=(1a10071)×(11a) \dfrac{1+a^{2n}}{1+a^{4n}} < \dfrac{1}{a^n} \\ \sum_{n=1}^{1007} \dfrac{1+a^{2n}}{1+a^{4n}} <\sum_{n=1}^{1007} \dfrac{1}{a^n} = \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{1-a} \right)

Similary n=110071+b2n1+b4n<(1b10071)×(11b)\sum_{n=1}^{1007} \dfrac{1+b^{2n}}{1+b^{4n}} < \left( \dfrac{1}{b^{1007}} - 1 \right) \times \left( \dfrac{1}{1-b} \right)

Multiplying the two, we get

n=110071+a2n1+a4n×n=110071+b2n1+b4n<(1a10071)×(1b10071)×1(1a)(1b) \sum_{n=1}^{1007} \dfrac{1+a^{2n}}{1+a^{4n}} \times \sum_{n=1}^{1007} \dfrac{1+b^{2n}}{1+b^{4n}} < \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) \times \dfrac{1}{\left(1-a \right) \left( 1-b \right)}

So, If we prove that (1a10071)×(1b10071)<1 \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) < 1, then we are done.

Since a2014+b2014=1a^{2014} + b^{2014} = 1 and 0a,b0 \leq a, b . It implies a,b<1a,b < 1. So, a1007>a2014a^{1007} > a^{2014} and b1007>b2014b^{1007} > b^{2014}. Which implies that a1007+b1007>a2014+b2014=1a^{1007} + b^{1007} > a^{2014} + b^{2014} = 1.

So,

1a1007b1007<0(1a1007)×(1b1007)<a1007×b1007(1a10071)×(1b10071)<11-a^{1007} - b^{1007} <0 \\ \left( 1- a^{1007} \right) \times \left(1- b^{1007} \right) < a^{1007} \times b^{1007} \\ \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) < 1

Therefore,

(n=11007x2n+1x4n+1)(n=11007y2n+1y4n+1)=(n=11007a2n+1a4n+1)(n=11007b2n+1b4n+1)<1(1a)(1b)=1(1x)(1y)\left( \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right) = \left( \sum_{n=1}^{1007} \dfrac{a^{2n}+1}{a^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{b^{2n}+1}{b^{4n}+1} \right) < \dfrac{1}{\left(1-a \right) \left( 1-b \right)} = \dfrac{1}{\left(1-|x| \right) \left( 1-|y| \right)}

Pheww!! Completed. \blacksquare.

Surya Prakash - 3 years, 8 months ago

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Thank you very much.

Nihar Mahajan - 3 years, 8 months ago

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Surya, explain the n=110071an \sum_{n=1}^{1007} \dfrac{1}{a^n} part.

Saarthak Marathe - 3 years, 8 months ago

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any hint

Dev Sharma - 3 years, 9 months ago

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hint is provided

Shivam Jadhav - 3 years, 9 months ago

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Shivam prove the below statement. If you do so, the problem is done.

Saarthak Marathe - 3 years, 9 months ago

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i want to look its solution..

Dev Sharma - 3 years, 9 months ago

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@Shivam Jadhav @Saarthak Marathe Please post a solution. My curiosity is running out and so is of some people.Thanks.

Nihar Mahajan - 3 years, 8 months ago

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Agree....

Dev Sharma - 3 years, 8 months ago

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I found another solution to this problem.This is my own solution. I did not know how to add pdf's to brilliant so I uploaded it on my google drive account. Please See the link below.

https://drive.google.com/file/d/0ByQHFlC74eP_dGxzOTBVaFZKTVE/view?usp=sharing

@Shivam Jadhav @Svatejas Shivakumar @Dev Sharma @Surya Prakash @Nihar Mahajan

Saarthak Marathe - 3 years, 7 months ago

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@Pi Han Goh

Shivam Jadhav - 3 years, 9 months ago

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If the following is proved then the problem is done, For x1x\le1,

Prove that 1x2n1+x2nx\frac{1-{x}^{2n}}{1+{x}^{2n}} \le x where nNn \in N

Saarthak Marathe - 3 years, 9 months ago

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@Shivam Jadhav Please post a solution.

Nihar Mahajan - 3 years, 8 months ago

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No not yet. Please wait for some more time. I think I am getting it.@Shivam Jadhav

Saarthak Marathe - 3 years, 8 months ago

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I think we have got too much time for this. Remember that we have only half an hour in RMO for a question.

Nihar Mahajan - 3 years, 8 months ago

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@Nihar Mahajan But once you get such a question by yourself then you will remember the method for your lifetime.

Saarthak Marathe - 3 years, 8 months ago

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@Saarthak Marathe Then , please post your solution. Too curious :P

Nihar Mahajan - 3 years, 8 months ago

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Sarthak wants to post the solution.

Shivam Jadhav - 3 years, 8 months ago

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@Shivam Jadhav Please post a solution.

Nihar Mahajan - 3 years, 8 months ago

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@Shivam Jadhav @Saarthak Marathe Please post a solution. Almost three weeks are over since this problem was posted and no one has got a solution so far.

And BTW, in the problem it should be mentioned that x and y \neq 1.

Brilliant Member - 3 years, 8 months ago

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It is understood.

Saarthak Marathe - 3 years, 8 months ago

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@Shivam Jadhav Post the solution.

Saarthak Marathe - 3 years, 8 months ago

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@Shivam Jadhav Please post the solution!!! I am losing my curiosity!!!!

Brilliant Member - 3 years, 8 months ago

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I think we are unnecessarily wasting our time asking him because he probably doesn't have any solution.

Nihar Mahajan - 3 years, 8 months ago

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@Nihar Mahajan Yes even I think so. Can we ask Calvin sir to help us in this problem (provided this problem actually exists :D)?

Brilliant Member - 3 years, 8 months ago

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@Shivam Jadhav Please refrain from mass tagging. Please try and limit it to five people. Thanks.

Sudeep Salgia - 3 years, 9 months ago

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