prove that $|x^y - y^x| > 2$ 
Prove that for any integers $x\neq y ,x,y>2, |x^y-y^x|>2$.

I know that the function $f(x) = \dfrac{\ln x}{x}$ is increasing for $x<e$ and decreasing for $x > e$. So $|x^y-y^x| = x^y - y^x \ge 1$ if $x < y$. I was thinking of proving something involving ratios (e.g. $f(x)/f(y)$ for $x < y$). I could possibly consider the case where x and y are consecutive and I could assume WLOG that $y>x$ so that it suffices to prove $x^y > y^x+2$ for $y>x>2.$ The inequality seems to be fairly weak; even for the smallest possible values of x and y, $y=4,x=3,$ we have $x^y = y^x+17.$ I think the sequence $\dfrac{x^{x+1}}{(x+1)^x} = x(1-\dfrac{1}{x+1})^x$  is an increasing function of x, and one could prove this using derivatives.

Source: A PuMAC 2008 problem.


Letting $f(x)=x^{1/x}$ then for integers $2<x<y,$ $$\begin{align}y^x-x^y&=f(x)^{xy}-f(y)^{xy}\\&=(f(y)-f(x))\sum_{k=0}^{xy-1} f(x)^k f(y)^{xy-1-k}\\&>xyf(y)^{xy-1}(f(y)-f(x)).\end{align}$$
the last step because $f$ is decreasing on $[3,\infty)$ and thus $f(x)>f(y).$
Now, $f(x)-f(y)=f'(c)(x-y)$ for some $c\in(x,y).$ And $$f'(x)=\frac{1-\ln x}{x^2}x^{1/x}=\frac{1-\ln x}{x^2}f(x).$$
Since $\frac{\ln x-1}{x^2}$ is decreasing and positive for $x\geq e^{3/2}<5$ and the same is true for $x^{1/x},$ the largest value for $|f'(c)|$ is $|f'(x)|$ if $x\geq 5.$
Adding that $f(y)^{xy-1}=\frac{y^x}{f(y)}$ you get:
$$\left|y^x-x^y\right|> (y-x)xy\cdot y^{x}\frac{\log x-1}{x^2}\frac{f(x)}{f(y)}>(y-x)y^x(\log x-1)$$
When $5\leq x<y$ are integers, $y-x\geq 1,$ $y^x>2$ and $\log x>\frac32.$ (You actually have $y^x\geq 6^5,$ so this would give you a lower bound of $3888$ when $x\geq 5.$)
So you are left with the cases $x=3$ and $x=4,$ which can be proved seperately. You can easily show that you only need to check a finite number of values $y$ in those cases, because $3^y>y^4$ for all but finitely many $y,$ and $4^y>y^5$ for all but finitely many $y.$

Looking at the graph for $f''(x),$ it seems like $|f'(x)|$ is decreasing for $x\geq 1.$ Above, I used a shorthand to avoid computing $f''$ directly, but if true, the rest of the argument works for $x=3$ and $4,$ too.
Then the lower bound for $x\geq 3$ is $(\log(x)-1)x^{x+1}>6.$
We still need $x\geq 3$ to get $f(x)>f(y).$
