Let \(x\) and \(y\) be real numbers such that \(x^2+y^2=1\) .Prove that \[\dfrac{1}{1+x^2}+ \dfrac{1}{1+y^2} + \dfrac{1}{1+xy} \geq \frac{3}{1+(\dfrac{x+y}{2})^2}\]

Please help me this as soon as possible.Thanks!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIMAGE

IMAGE

Log in to reply

Great! A solution through classical inequalities should be better.

Log in to reply

I Prefer Trigonometry . Also i tried to use cauchy, titu but couldn't succeed therefore i switched to trigonometric substitution

Log in to reply

@rohit kumar @Aniket Sanghi @ARYAN GOYAT @Archit Agrawal @Aditya Chauhan

Can anyone post an algebraic proof?

Log in to reply

@Anik Mandal @Harsh Shrivastava @Sharky Kesa

How's the proof?

Log in to reply

hey @Anik Mandal i got the result!

Log in to reply

Looks like an application of Cauchy's/Titu's

Log in to reply

Did you get the result?

Log in to reply

@Sharky Kesa

Log in to reply

Setting xy=t maybe fruitful, but I have not tried it.

Log in to reply

Are you sure the problem is correct, because x=0.5 and y=\(-0.5\) are not satisfying the condition.

Log in to reply

They are not satisfying first condition only.

Log in to reply

Oh sorry I misread x^2 +y^2 as x+y.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@Harsh Shrivastava Can you help?

Log in to reply

First I would like to prove the following: \(\dfrac{1}{1+x^2} + \dfrac{1}{1+y^2} \geq \dfrac{2}{1+{\frac{(x+y)}{2}}^{2}} \dots (1) \) for \( \dfrac{1}{4} \leq xy \leq \dfrac{1}{2} \)

\( \dfrac{2 + x^2 + y^2}{1 + x^2 + y^2 + {x}^{2}{y}^{2}} \geq \dfrac{2}{1 + \frac{1+2xy}{4}} \)

\( \dfrac{3}{2+{x}^{2}{y}^{2}} \geq \dfrac{2}{\frac{5}{4} + \frac{xy}{2}} \)

\( 8{xy}^{2} -6xy +1 \leq 0 \)

This is true for \( \frac{1}{4} \leq xy \leq \frac{1}{2} \). Thus it has been proved.

\( \dfrac{1}{1+xy} \geq \dfrac{1}{1+{\frac{(x+y)}{4}}^{2}} \dots (2) \) (trivial)

Now add (1) and (2).Thus the result is proved for \( \dfrac{1}{4} \leq xy \leq \dfrac{1}{2} \).

What remains (\( 0 \leq xy \leq \frac{1}{4} \)) is simple.

The minimum value of the L.H.S in this interval is \( \dfrac{16}{11} + \dfrac{5}{4} \). The maximum value of the R.H.S is \( \dfrac{12}{5} \).

This proves the result for \( 0 \leq xy \leq \dfrac{1}{2} \).

Log in to reply

I Have sent you that integral on mail. check it out!

Log in to reply

Very Slick! +1. i was expecting an algebraic solution from you

Log in to reply

thanks !. nice use of trig substitution by the way.

Log in to reply

actually I misread the question.fhe proof Only Works For Positive x And y.

Log in to reply

Ohh!. i didn't noticed. BTW Thanks! :)

Log in to reply

(Sorry for asking this at wrong place.)

Log in to reply

But Last year KVPY Math and chem were quite easy. Bio was a nightmare(atleast for me)

My physics didn't went that well as it had some problems from optics which wasn't taught at that point of time.

But yeah if you are appearing for NSEA Then do master Gravitation!!.

For KVPY Do study current electricity , optics in physics .

Hydrocarbons in chemistry and basics taught in class 10th (Mensuration , Volume and surface area , Elementary Number theory) that will be enought

And Yeah Geometry is also important.

All the best! :)

Log in to reply

Any further tips?

Thanks.

Log in to reply

chem is easy but they ask some questions of organic chemistry which i suppose is there in phase-3.

Studying NCERT Will be enough for that

physics will be easy of the topics you have been taught.

All the best!

Log in to reply

Log in to reply

@rohit kumar also.he also qualified it

You can askLog in to reply

Log in to reply