(x^2 + y^2)/xy > 2

Jazz10 · #1 $Old$ November 3rd, 2009, 07:19 PM

(x^2 + y^2)/xy = x/y + y/x > 2. I can clearly see why this is true, but can't find a proof, need this as a lemma to a harder problem. Any help is appreciated.

alexmahone · #2 $Old$ November 3rd, 2009, 07:45 PM

Quote:

Originally Posted by Jazz10 $View Post$

(x^2 + y^2)/xy = x/y + y/x > 2. I can clearly see why this is true, but can't find a proof, need this as a lemma to a harder problem. Any help is appreciated.

$\frac{\frac{x}{y}+\frac{y}{x}}2\geq\sqrt{\frac{x}{y}\frac{y}{x}}$ (By AM-GM inequality)

$\frac{x}{y}+\frac{y}{x}\geq2$

artvandalay11 · #3 $Old$ November 3rd, 2009, 07:49 PM

Quote:

Originally Posted by Jazz10 $View Post$

(x^2 + y^2)/xy = x/y + y/x > 2. I can clearly see why this is true, but can't find a proof, need this as a lemma to a harder problem. Any help is appreciated.

I think you meant greater than or equal to, otherwise it's wrong

for ex. x=y=1

chisigma · #4 $Old$ November 4th, 2009, 06:07 AM

Given the function...

$f(x,y)= \frac{xy}{x^{2} + y^{2}}$ (1)

... it can be written in polar form with the substitution of variables...

$x=\rho\cdot \cos \theta$

$y=\rho\cdot \sin \theta$ (2)

... so that it becomes...

$f(\rho,\theta)= \frac{\rho^{2}\cdot \sin \theta\cdot \cos \theta}{\rho^{2}} = \sin \theta\cdot \cos \theta$ (3)

The (3) shows that $f(*,*)$ in effects is function of the $\theta$ only and its maxima [or minima...] are located on the $\theta$ for which is...

$cos^{2} \theta = \sin^{2} \theta \rightarrow \theta = \frac{k}{4}\cdot \pi$ (4)

... where is...

$|f(\theta)|=\frac{1}{2}$ (5)

From (5) we have immediately...

$|f(\theta)|\le \frac{1}{2} \rightarrow \frac{1}{|f(\theta)|} \ge 2$ (6)

Kind regards

$\chi$ $\sigma$

CaptainBlack · #5 $Old$ November 4th, 2009, 06:45 AM

Quote:

Originally Posted by Jazz10 $View Post$

(x^2 + y^2)/xy = x/y + y/x > 2. I can clearly see why this is true, but can't find a proof, need this as a lemma to a harder problem. Any help is appreciated.

Assume $xy >0$

$x^2-2xy+y^2=(x-y)^2 \ge 0$

hence:

$x^2+y^2 \ge 2xy$

so:

$\frac{x^2+y^2}{xy} \ge 2$

Note1 you need the assumption that $xy>0$ , as you can see by putting $x=2, y=-1$

Note2 you need $\ge$ rather than $>$ since when $x=y=1$ : $\frac{x^2+y^2}{xy}=2$

CB