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If x,yx,yx,y are real numbers such that x2014+y2014=1x^{2014}+y^{2014}=1x2014+y2014=1 then prove that:
(∑n=11007x2n+1x4n+1)(∑n=11007y2n+1y4n+1)<1(1−x)(1−y)\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)}(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)<(1−x)(1−y)1
Note by Shivam Jadhav 4 years, 4 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=-1x=−1 and y=0y=0y=0. Then we get
∑n=11007x2n+1x4n+1=1007∑n=11007y2n+1y4n+1=10071(1−x)(1−y)=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} n=1∑1007x4n+1x2n+1=1007n=1∑1007y4n+1y2n+1=1007(1−x)(1−y)1=21
So, according to our question it forces that 1007×1007<1/21007 \times 1007 < 1/21007×1007<1/2, which is impossible.
The question should be actually like this:
If xxx, yyyare real numbers such that x2014+y2014=1x^{2014}+y^{2014}=1x2014+y2014=1 then prove that:
(∑n=11007x2n+1x4n+1)(∑n=11007y2n+1y4n+1)<1(1−∣x∣)(1−∣y∣)\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)}(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)<(1−∣x∣)(1−∣y∣)1
PROOF:
Let ∣x∣=a|x| = a∣x∣=a and ∣y∣=b|y| = b∣y∣=b. So, the given constraint changes into a2014+b2014=1a^{2014} + b^{2014} = 1a2014+b2014=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)a4n+1>2(a2n+1)2=21×(a2n+1)×(a2n+1)>an(a2n+1). Which implies that
1+a2n1+a4n<1an∑n=110071+a2n1+a4n<∑n=110071an=(1a1007−1)×(11−a) \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)1+a4n1+a2n<an1n=1∑10071+a4n1+a2n<n=1∑1007an1=(a10071−1)×(1−a1)
Similary ∑n=110071+b2n1+b4n<(1b1007−1)×(11−b)\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)∑n=110071+b4n1+b2n<(b10071−1)×(1−b1)
Multiplying the two, we get
∑n=110071+a2n1+a4n×∑n=110071+b2n1+b4n<(1a1007−1)×(1b1007−1)×1(1−a)(1−b) \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)}n=1∑10071+a4n1+a2n×n=1∑10071+b4n1+b2n<(a10071−1)×(b10071−1)×(1−a)(1−b)1
So, If we prove that (1a1007−1)×(1b1007−1)<1 \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) < 1(a10071−1)×(b10071−1)<1, then we are done.
Since a2014+b2014=1a^{2014} + b^{2014} = 1 a2014+b2014=1 and 0≤a,b0 \leq a, b 0≤a,b. It implies a,b<1a,b < 1a,b<1. So, a1007>a2014a^{1007} > a^{2014}a1007>a2014 and b1007>b2014b^{1007} > b^{2014}b1007>b2014. Which implies that a1007+b1007>a2014+b2014=1a^{1007} + b^{1007} > a^{2014} + b^{2014} = 1a1007+b1007>a2014+b2014=1.
So,
1−a1007−b1007<0(1−a1007)×(1−b1007)<a1007×b1007(1a1007−1)×(1b1007−1)<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 1−a1007−b1007<0(1−a1007)×(1−b1007)<a1007×b1007(a10071−1)×(b10071−1)<1
Therefore,
(∑n=11007x2n+1x4n+1)(∑n=11007y2n+1y4n+1)=(∑n=11007a2n+1a4n+1)(∑n=11007b2n+1b4n+1)<1(1−a)(1−b)=1(1−∣x∣)(1−∣y∣)\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)}(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)=(n=1∑1007a4n+1a2n+1)(n=1∑1007b4n+1b2n+1)<(1−a)(1−b)1=(1−∣x∣)(1−∣y∣)1
Pheww!! Completed. ■\blacksquare■.
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Thank you very much.
Surya, explain the ∑n=110071an \sum_{n=1}^{1007} \dfrac{1}{a^n} ∑n=11007an1 part.
any hint
hint is provided
Shivam prove the below statement. If you do so, the problem is done.
i want to look its solution..
@Shivam Jadhav @Saarthak Marathe Please post a solution. My curiosity is running out and so is of some people.Thanks.
Agree....
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
@Pi Han Goh
If the following is proved then the problem is done, For x≤1x\le1x≤1,
Prove that 1−x2n1+x2n≤x\frac{1-{x}^{2n}}{1+{x}^{2n}} \le x 1+x2n1−x2n≤x where n∈Nn \in N n∈N
@Shivam Jadhav Please post a solution.
No not yet. Please wait for some more time. I think I am getting it.@Shivam Jadhav
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 – But once you get such a question by yourself then you will remember the method for your lifetime.
@Saarthak Marathe – Then , please post your solution. Too curious :P
Sarthak wants to post the solution.
@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.
It is understood.
@Shivam Jadhav Post the solution.
@Shivam Jadhav Please post the solution!!! I am losing my curiosity!!!!
I think we are unnecessarily wasting our time asking him because he probably doesn't have any solution.
@Nihar Mahajan Yes even I think so. Can we ask Calvin sir to help us in this problem (provided this problem actually exists :D)?
@Nihar Mahajan @Calvin Lin @Daniel Liu @Joel Tan @Harsh Shrivastava @Xuming Liang @Ammar Fathin Sabili @Satyajit Mohanty
@Shivam Jadhav Please refrain from mass tagging. Please try and limit it to five people. Thanks.
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Top NewestThere 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
n=1∑1007x4n+1x2n+1=1007n=1∑1007y4n+1y2n+1=1007(1−x)(1−y)1=21
So, according to our question it forces that 1007×1007<1/2, which is impossible.
The question should be actually like this:
If x, yare real numbers such that x2014+y2014=1 then prove that:
(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)<(1−∣x∣)(1−∣y∣)1
PROOF:
Let ∣x∣=a and ∣y∣=b. So, the given constraint changes into a2014+b2014=1.
Since, a4n+1>2(a2n+1)2=21×(a2n+1)×(a2n+1)>an(a2n+1). Which implies that
1+a4n1+a2n<an1n=1∑10071+a4n1+a2n<n=1∑1007an1=(a10071−1)×(1−a1)
Similary ∑n=110071+b4n1+b2n<(b10071−1)×(1−b1)
Multiplying the two, we get
n=1∑10071+a4n1+a2n×n=1∑10071+b4n1+b2n<(a10071−1)×(b10071−1)×(1−a)(1−b)1
So, If we prove that (a10071−1)×(b10071−1)<1, then we are done.
Since a2014+b2014=1 and 0≤a,b. It implies a,b<1. So, a1007>a2014 and b1007>b2014. Which implies that a1007+b1007>a2014+b2014=1.
So,
1−a1007−b1007<0(1−a1007)×(1−b1007)<a1007×b1007(a10071−1)×(b10071−1)<1
Therefore,
(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)=(n=1∑1007a4n+1a2n+1)(n=1∑1007b4n+1b2n+1)<(1−a)(1−b)1=(1−∣x∣)(1−∣y∣)1
Pheww!! Completed. ■.
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Thank you very much.
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Surya, explain the ∑n=11007an1 part.
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any hint
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hint is provided
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Shivam prove the below statement. If you do so, the problem is done.
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i want to look its solution..
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@Shivam Jadhav @Saarthak Marathe Please post a solution. My curiosity is running out and so is of some people.Thanks.
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Agree....
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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
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@Pi Han Goh
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If the following is proved then the problem is done, For x≤1,
Prove that 1+x2n1−x2n≤x where n∈N
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@Shivam Jadhav Please post a solution.
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No not yet. Please wait for some more time. I think I am getting it.@Shivam Jadhav
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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.
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But once you get such a question by yourself then you will remember the method for your lifetime.
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Then , please post your solution. Too curious :P
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Sarthak wants to post the solution.
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@Shivam Jadhav Please post a solution.
Log in to reply
@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 = 1.
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It is understood.
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@Shivam Jadhav Post the solution.
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@Shivam Jadhav Please post the solution!!! I am losing my curiosity!!!!
Log in to reply
I think we are unnecessarily wasting our time asking him because he probably doesn't have any solution.
Log in to reply
@Nihar Mahajan Yes even I think so. Can we ask Calvin sir to help us in this problem (provided this problem actually exists :D)?
Log in to reply
@Nihar Mahajan @Calvin Lin @Daniel Liu @Joel Tan @Harsh Shrivastava @Xuming Liang @Ammar Fathin Sabili @Satyajit Mohanty
Log in to reply
@Shivam Jadhav Please refrain from mass tagging. Please try and limit it to five people. Thanks.
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