For $x+y=n$, $y^x < x^y$ if $x(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?
 A: 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 $$ 1<x<y<e,\ x,y\neq 1\ \Rightarrow y^x>x^y.$$  
A: The claim is false. $1^{10}<10^1$. Even if you exclude the $x=1$ case, $2^3<3^2$.
A: 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.
