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How can we proof following?

if $X < Y$, then:

$X^{Y} > Y^{X}$ ,

Where X, and Y are integers. Also $X,Y > 1$.

Except a special case $2^{3} < 3^{2}$.

I think for other variables $X,Y$ above equation is correct.

I need to for pumping a lemma question

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Is it well known and we can leave at this stage in answer –  Grijesh Chauhan Jan 1 '13 at 6:27
The title and question have the x < y condition backward, did you mean it to be like that? –  Amzoti Jan 1 '13 at 6:33
Take logarithms of the two sides and use Lagrange's mean value theorem. –  Amihai Zivan Jan 1 '13 at 6:36
this is a problem iit aspirants always have at their fingertips. –  Koushik Jan 1 '13 at 7:11
Also $2 < 4$ but $2^4 = 4^2$. –  anthus Jan 1 '13 at 7:15

2 Answers 2

up vote 2 down vote accepted

Let $x<y$, so $x^y>y^x$.

Since $\ln \alpha$ is monotonic at $(0,\infty)$ it is enough to show that $\ln x^y > \ln y^x$ which translates to $y\ln x> x\ln y \iff \frac{\ln x}{x}>\frac{\ln y}{y}$. Define $f(\alpha)=\frac {\ln \alpha}{\alpha}$, we need to prove that $f(x)>f(y) $ for $ x<y $.

Investagiting $f$ would yield that it has a global maximum at $\alpha=e$ and it's monotone decreasing afterwards, concluding that for $y>x>e$, $f(y)<f(x)$, as needed.

My comment about using mean value theorem was a mistake, sorry about that.

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Thanks Amihai Zivan!! .. –  Grijesh Chauhan Jan 11 '13 at 7:01

Study the behavior of $~\large \log(x^{1/x})~$ = $~\large \frac{\log(x)}{x}$.

Use the derivative = $0$ at $e$, and derivative $<0~$ for $~x>e$, so $\large x^{1/x}$ is decreasing.

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thanks K.Ghosh! –  Grijesh Chauhan Jan 1 '13 at 7:14
you are welcome –  Koushik Jan 1 '13 at 7:15
@Amzoti You messed up the math in that edit there... –  TMM Jan 1 '13 at 13:13
@TMM, thanks for catching and correcting! –  Amzoti Jan 1 '13 at 14:17

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