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(updated) I'd like to use this property for my research, but it's somewhat messy to prove.

$$\text{For all natural number $x,y$ such that $x+y=n$ and $1<x<y<n$, then $y^x < x^y$}.$$

For example, let $x=3, y=7$. Then $y^x = y^3 = 343$ and $x^y = 3^7 = 2187$. Any suggestion on how to prove this?

Federico Magallanez
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    Is $n$ a positive integer? Is there any bound on $n$? This is not true though. $n=6=2+4$ and $2^4 = 4^2$. – Aryabhata Feb 24 '12 at 16:12
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    Well that is not really true now is it... $(-4)^2>2^{-4}$ for example. An even better example is $3^2>2^3$... Having said that, if both $x,y>e=2.7818\dots$, then we always have the implication $x<y\Rightarrow y^x<x^y$. – Eric Naslund Feb 24 '12 at 16:13
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    @Fed: The counterexample I gave still works for(or rather, against) your update. – Aryabhata Feb 24 '12 at 16:20
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    $x=2,y=3,n=5$ still stands as a counter example. – Inquest Feb 24 '12 at 16:21
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    It is true if $x$ and $y$ are integers and $n \gt 6$ – Henry Feb 24 '12 at 16:21

3 Answers3

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Lets ask the general question

When does $x<y\Rightarrow y^x<x^y$ hold for real $x,y$?

Taking logarithms, we are asking when $x<y$ implies $x\log y<y\log x$, or when $$\frac{x}{\log x}<\frac{y}{\log y}.$$ Notice this is the same as when the function $f(x)=\frac{x}{\log x}$ is increasing. To answer that we look at the derivative, and since

$$f^{'}(x)=\frac{\log(x)-1}{\log^2(x)}$$

we see the derivative is only positive, and that the function is only increasing, when $x>e$. Hence if $x,y>e$ you will have $$x<y\Rightarrow y^x<x^y.$$

If $0<x,y<e$ and $x,y\neq 1$, then the derivative is actually negative, and we get the opposite $$ 1<x<y<e,\ x,y\neq 1\ \Rightarrow y^x>x^y.$$

Eric Naslund
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The claim is false. $1^{10}<10^1$. Even if you exclude the $x=1$ case, $2^3<3^2$.

Chris Eagle
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I proved this in the special case $x = 99, y = 100$, here. As others have pointed out, what you really want to hold is the following:

Statement: Let $x, y \in \mathbb{R}$. Then $y > x > e$ implies $x^y > y^x$.

Proof:. Write $y = x + z$, where $z > 0$. Then,

$$\begin{align} x^y > y^x &\iff x^x x^z > y^x \\ &\iff x^z > \left(\frac{x+z}{x} \right)^x \\ &\iff x^z > \left( 1 + \frac{z}{x} \right)^x. \end{align}$$

The right hand side $\left(1 + \frac{z}{x} \right)^x$ is monotone increasing with limit $e^z$. Since the left hand size is strictly greater than $e^z$ (as $x > e$), it follows that the inequality always holds.

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JavaMan
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