Exercise 1.10 [Section 1, BOMC-2011]

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Labib
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Exercise 1.10 [Section 1, BOMC-2011]

Unread post by Labib » Wed Oct 26, 2011 12:43 pm

If $x,y> 0$, then prove that $\sqrt{\frac{x^{2}}{y}} + \sqrt{\frac{y^{2}}{x}}\geqslant \sqrt{x}+\sqrt{y}$
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by Labib » Wed Oct 26, 2011 12:44 pm

TOWFIQUL wrote:Anybody solved the exercise 1.10?
mugdho.snigdho wrote:Could u plz state the problem 1.10 (i'm in the class & apart from the book right now ) ?
nafistiham wrote:unfortunately, i couldn't solve this problem $1.10$ .again stating the problem for mugdho.snigdho and myself.hoping to get some hint.
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by Labib » Wed Oct 26, 2011 12:45 pm

iPavel wrote:@toweikul
Assume without loss of genarility,$x\geq y$, then $\sqrt {x}\geq \sqrt {y}$.

Now consider two set,$A=\left \{x,y\right \}$ and $B=\left \{\sqrt {x},\sqrt {y}\right \}$

Using rearrangement inequality,$x\sqrt {x}+y\sqrt {y}\geq x\sqrt {y}+y\sqrt {x}=\sqrt {xy}(\sqrt {x}+\sqrt {y})$
Last edited by Labib on Wed Oct 26, 2011 1:04 pm, edited 1 time in total.
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by Labib » Wed Oct 26, 2011 12:46 pm

Labib wrote:Tiham,
Would you like some hints?
Here are some if you want...
(a) Try squaring the sides and then simplify...
It should come to this-
It is enough to proof that,
$x^3+y^3\geq xy(x+y)$

(b) Now, without loss of generality, assume $x\geq y$,

(c) Now, substituting $x$ with $y+k$ follows the result.
The equality holds when $k=0$.
Hasib wrote:To solve the proble 1.10, you have to define two case while $x \geq y$ and $x<y$ and u can try by squaring two side. :D
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by *Mahi* » Wed Oct 26, 2011 12:54 pm

The given equation is symmetric with respect to $x,y$ and so one case would be enough.
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by Hasib » Wed Oct 26, 2011 1:00 pm

*Mahi* wrote:The given equation is symmetric with respect to $x,y$ and so one case would be enough.
as you wish, but i was slapped when it was my first approach. I go by backward, and discovered that it have to be $x>y$. But, i skip that i divide $x^2-y^2$ that means it must be $x>y$. And if i divide by $x^2-y^2$ and change the direction, it must be $x<y$, thus proved....
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by photon » Wed Oct 26, 2011 1:00 pm

the equality holds if $x=y$.let's assume, $x>y$
$\Rightarrow \frac{\sqrt{x}}{\sqrt{y}}>\frac{\sqrt{y}}{\sqrt{x}}$
$\Rightarrow \sqrt{x}\frac{\sqrt{x}-\sqrt{y}}{\sqrt{y}}>\sqrt{y}\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}$
$\Rightarrow \frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}>\sqrt{x}+\sqrt{y}$
$\Rightarrow \sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}>\sqrt{x}+\sqrt{y}$
done.
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by *Mahi* » Wed Oct 26, 2011 1:26 pm

To avoid the cluster of root signs , you may substitute $x,y$ with $a^2,b^2$ as it is given that $x,y > 0$
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by sourav das » Wed Oct 26, 2011 1:48 pm

$( \sqrt{x}+ \sqrt{y} ) \left ( \sqrt{x}-\sqrt{y} \right )^2\geq 0$ and simplify it to:
$\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}\geq \sqrt{x}+\sqrt{y}$
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Re: Exercise 1.10 [Section 1, BOMC-2011]

Unread post by sm.joty » Wed Oct 26, 2011 2:27 pm

আমার মনে হয় সবচে সহজ পদ্ধতি হল Contradiction . যেহেতু x,y>0 তাই \[\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\]
এখন যা প্রমান করতে বলা হয়েছে টার বিপরিতটা সঠিক ধরে আগালেই হয়।
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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