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
4 years, 1 month 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. $\blacksquare$.

- 3 years, 12 months ago

Thank you very much.

- 3 years, 12 months ago

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

- 3 years, 12 months ago

any hint

- 4 years, 1 month ago

hint is provided

- 4 years, 1 month ago

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

- 4 years, 1 month ago

i want to look its solution..

- 4 years, 1 month ago

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

- 4 years ago

Agree....

- 4 years ago

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.

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

- 3 years, 11 months ago

- 4 years, 1 month ago

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 $n \in N$

- 4 years, 1 month ago

- 4 years ago

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

- 4 years ago

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

- 4 years ago

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

- 4 years ago

Then , please post your solution. Too curious :P

- 4 years ago

Sarthak wants to post the solution.

- 4 years ago

- 4 years ago

@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.

- 4 years ago

@Shivam Jadhav Post the solution.

- 4 years ago

@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.

- 4 years ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago