camp exam problem-9

Discussion on Bangladesh National Math Camp
User avatar
nafistiham
Posts:829
Joined:Mon Oct 17, 2011 3:56 pm
Location:24.758613,90.400161
Contact:
camp exam problem-9

Unread post by nafistiham » Fri Nov 04, 2011 10:58 pm

Let $0 \leq x,y \leq $1 . Prove that,


\[\frac{1}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+y^2}} \leq \frac{2}{\sqrt{1+xy}}\]



[Hint : any positive real number can be written as $e^{u}$ for some real number $u$ . You can use this fact without proof. If you can solve the problem without using this hint, then you will get bonus point]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

User avatar
nafistiham
Posts:829
Joined:Mon Oct 17, 2011 3:56 pm
Location:24.758613,90.400161
Contact:

Re: camp exam problem-9

Unread post by nafistiham » Sat Nov 05, 2011 1:10 am

does this $e$ means the exponential?
if it is could someone show me the proof of the hint?
i really couldn't understand the whole thing [hint]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

User avatar
sm.joty
Posts:327
Joined:Thu Aug 18, 2011 12:42 am
Location:Dhaka

Re: camp exam problem-9

Unread post by sm.joty » Sat Nov 05, 2011 10:57 am

I think I can solve it. But I always confused about my solution. :?

And here $e$ is Eular's constant.As a student of secondary (I think), you already know about it. Otherwise follow this link
http://en.wikipedia.org/wiki/E_%28mathe ... onstant%29
It is clear that the equality holds if and only if $x=y$
now it's enough to prove it for the other case.
According to the hints we can write,
$e^a=\sqrt{1+x^2}$ ,
$e^b=\sqrt{1+y^2}$,
$e^c=\sqrt{1+xy}$
Now, clearly $xy>0$
$\Rightarrow 1+xy>1$
$\Rightarrow e^c>1$.................(1)
Again, $1+x^2<2$
here both side of the inequality is positive. So we can apply squire root,
$\sqrt{1+x^2}<\sqrt{2}$
$\Rightarrow e^a<\sqrt{2}$
so $\Rightarrow e^b<\sqrt{2}$
So, $e^{a+b}<2$
$\Rightarrow \frac{1}{e^{a+b}}>\frac{1}{2}$...................(2)
Again, $e^{a}+e^{b}<2\sqrt{2}$
$\Rightarrow -(e^{a}+e^{b})>-2\sqrt{2}$........................(3)

Multiplying (2) and (3) we get,
$-(\frac{e^a+e^b}{e^{a+b}})>-\sqrt{2}$..........................(4)

Multiplying (1) and (4) we get,
$-e^c(\frac{e^a+e^b}{e^{a+b}})>-\sqrt{2}$
$\Rightarrow e^c(\frac{e^a+e^b}{e^{a+b}})<\sqrt{2}$

SO, $ e^c(\frac{e^a+e^b}{e^{a+b}})\leq \sqrt{2}$
Is there any error ? :?:
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

Post Reply