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!

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...

## Comments

Sort by:

TopNewestIMAGE

IMAGE

Log in to reply

– Anik Mandal · 7 months ago

Great! A solution through classical inequalities should be better.Log in to reply

– Prakhar Bindal · 7 months ago

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

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

Can anyone post an algebraic proof? – Prakhar Bindal · 7 months ago

Log in to reply

@Anik Mandal @Harsh Shrivastava @Sharky Kesa

How's the proof? – Prakhar Bindal · 7 months ago

Log in to reply

hey @Anik Mandal i got the result! – Prakhar Bindal · 7 months ago

Log in to reply

Looks like an application of Cauchy's/Titu's – Sharky Kesa · 7 months, 1 week ago

Log in to reply

– Anik Mandal · 7 months, 1 week ago

Did you get the result?Log in to reply

@Sharky Kesa – Anik Mandal · 7 months, 1 week ago

Log in to reply

Setting xy=t maybe fruitful, but I have not tried it. – Harsh Shrivastava · 7 months, 1 week ago

Log in to reply

Are you sure the problem is correct, because x=0.5 and y=\(-0.5\) are not satisfying the condition. – Harsh Shrivastava · 7 months, 1 week ago

Log in to reply

– Anik Mandal · 7 months, 1 week ago

They are not satisfying first condition only.Log in to reply

– Harsh Shrivastava · 7 months, 1 week ago

Oh sorry I misread x^2 +y^2 as x+y.Log in to reply

– Anik Mandal · 7 months, 1 week ago

I believe this was an RMO problem of some year.Anyways you'll have your RMO this Sunday right?Log in to reply

– Harsh Shrivastava · 7 months, 1 week ago

Bro I'll try this problem tonight because I have to goto Fiitjee after some time, and need to study chemistry, The problem seems to be tricky.Log in to reply

– Anik Mandal · 7 months, 1 week ago

Ok!Are you there on Slack or hangouts?I mean are you active?Log in to reply

– Harsh Shrivastava · 7 months, 1 week ago

Yeah, wbu?Log in to reply

– Anik Mandal · 7 months, 1 week ago

Same.Log in to reply

@Harsh Shrivastava Can you help? – Anik Mandal · 7 months, 1 week ago

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 \( 0 \leq xy \leq \dfrac{1}{2} \). – Rohit Kumar · 7 months ago

Log in to reply

– Prakhar Bindal · 6 months, 3 weeks ago

I Have sent you that integral on mail. check it out!Log in to reply

– Prakhar Bindal · 7 months ago

Very Slick! +1. i was expecting an algebraic solution from youLog in to reply

– Rohit Kumar · 7 months ago

thanks !. nice use of trig substitution by the way.Log in to reply

– Rohit Kumar · 7 months ago

actually I misread the question.fhe proof Only Works For Positive x And y.Log in to reply

– Prakhar Bindal · 6 months, 4 weeks ago

Ohh!. i didn't noticed. BTW Thanks! :)Log in to reply

(Sorry for asking this at wrong place.) – Harsh Shrivastava · 6 months, 4 weeks ago

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! :) – Prakhar Bindal · 6 months, 4 weeks ago

Log in to reply

Any further tips?

Thanks. – Harsh Shrivastava · 6 months, 4 weeks ago

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! – Prakhar Bindal · 6 months, 4 weeks ago

Log in to reply

– Harsh Shrivastava · 6 months, 4 weeks ago

Thanks for the tip.Log in to reply

@rohit kumar also.he also qualified it – Prakhar Bindal · 6 months, 4 weeks ago

You can askLog in to reply

– Prakhar Bindal · 6 months, 4 weeks ago

My pleasure!. do well :)Log in to reply