pow$(X,Y)$ $>$ pow$(Y,X)$, if $X<Y$.

Question

How can we proof following?

if $X < Y$, then:

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

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

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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}

Answer 1

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.

Answer 2

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.