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Inequality proof problem of the day # 1

If \(x,y\) are real numbers such that \(x^{2014}+y^{2014}=1\) then prove that:

\[\left(\sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left(\sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right)< \dfrac{1}{(1-x)(1-y)}\]

Note by Shivam Jadhav
1 year, 10 months ago

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

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

\[ \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} = 1007 \\ \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} = 1007 \\ \dfrac{1}{\left(1-x \right) \left(1-y \right)} = \dfrac{1}{2} \]

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


The question should be actually like this:

If \(x\), \(y\)are real numbers such that \(x^{2014}+y^{2014}=1\) then prove that:

\[\left( \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right) < \dfrac{1}{\left(1-|x| \right) \left( 1-|y| \right)}\]


PROOF:

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

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

\[ \dfrac{1+a^{2n}}{1+a^{4n}} < \dfrac{1}{a^n} \\ \sum_{n=1}^{1007} \dfrac{1+a^{2n}}{1+a^{4n}} <\sum_{n=1}^{1007} \dfrac{1}{a^n} = \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{1-a} \right)\]

Similary \(\sum_{n=1}^{1007} \dfrac{1+b^{2n}}{1+b^{4n}} < \left( \dfrac{1}{b^{1007}} - 1 \right) \times \left( \dfrac{1}{1-b} \right)\)

Multiplying the two, we get

\[ \sum_{n=1}^{1007} \dfrac{1+a^{2n}}{1+a^{4n}} \times \sum_{n=1}^{1007} \dfrac{1+b^{2n}}{1+b^{4n}} < \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) \times \dfrac{1}{\left(1-a \right) \left( 1-b \right)}\]

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

Since \(a^{2014} + b^{2014} = 1 \) and \(0 \leq a, b \). It implies \(a,b < 1\). So, \(a^{1007} > a^{2014}\) and \(b^{1007} > b^{2014}\). Which implies that \(a^{1007} + b^{1007} > a^{2014} + b^{2014} = 1\).

So,

\[1-a^{1007} - b^{1007} <0 \\ \left( 1- a^{1007} \right) \times \left(1- b^{1007} \right) < a^{1007} \times b^{1007} \\ \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) < 1 \]

Therefore,

\[\left( \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right) = \left( \sum_{n=1}^{1007} \dfrac{a^{2n}+1}{a^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{b^{2n}+1}{b^{4n}+1} \right) < \dfrac{1}{\left(1-a \right) \left( 1-b \right)} = \dfrac{1}{\left(1-|x| \right) \left( 1-|y| \right)}\]

Pheww!! Completed. \(\blacksquare\). Surya Prakash · 1 year, 9 months ago

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@Surya Prakash Surya, explain the \( \sum_{n=1}^{1007} \dfrac{1}{a^n} \) part. Saarthak Marathe · 1 year, 9 months ago

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@Surya Prakash Thank you very much. Nihar Mahajan · 1 year, 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 · 1 year, 9 months ago

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@Nihar Mahajan Agree.... Dev Sharma · 1 year, 9 months ago

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any hint Dev Sharma · 1 year, 10 months ago

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@Dev Sharma hint is provided Shivam Jadhav · 1 year, 10 months ago

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@Shivam Jadhav i want to look its solution.. Dev Sharma · 1 year, 10 months ago

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@Shivam Jadhav Shivam prove the below statement. If you do so, the problem is done. Saarthak Marathe · 1 year, 10 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 · 1 year, 8 months ago

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@Shivam Jadhav Please post the solution!!! I am losing my curiosity!!!! Svatejas Shivakumar · 1 year, 9 months ago

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@Svatejas Shivakumar I think we are unnecessarily wasting our time asking him because he probably doesn't have any solution. Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan @Nihar Mahajan Yes even I think so. Can we ask Calvin sir to help us in this problem (provided this problem actually exists :D)? Svatejas Shivakumar · 1 year, 9 months ago

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@Shivam Jadhav Post the solution. Saarthak Marathe · 1 year, 9 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. Svatejas Shivakumar · 1 year, 9 months ago

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@Svatejas Shivakumar It is understood. Saarthak Marathe · 1 year, 9 months ago

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@Shivam Jadhav Please post a solution. Nihar Mahajan · 1 year, 9 months ago

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@Shivam Jadhav Please post a solution. Nihar Mahajan · 1 year, 9 months ago

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@Nihar Mahajan Sarthak wants to post the solution. Shivam Jadhav · 1 year, 9 months ago

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@Nihar Mahajan No not yet. Please wait for some more time. I think I am getting it.@Shivam Jadhav Saarthak Marathe · 1 year, 9 months ago

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@Saarthak Marathe 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 · 1 year, 9 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 · 1 year, 9 months ago

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@Saarthak Marathe Then , please post your solution. Too curious :P Nihar Mahajan · 1 year, 9 months ago

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If the following is proved then the problem is done, For \(x\le1\),

Prove that \(\frac{1-{x}^{2n}}{1+{x}^{2n}} \le x \) where \(n \in N \) Saarthak Marathe · 1 year, 10 months ago

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@Pi Han Goh Shivam Jadhav · 1 year, 10 months ago

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@Shivam Jadhav @Shivam Jadhav Please refrain from mass tagging. Please try and limit it to five people. Thanks. Sudeep Salgia · 1 year, 10 months ago

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