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.

 A: 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 $f(x)>f(y).$
A: Let us prove that $|x^y - y^x| \ge 17$.
WLOG, assume that $3 \le x < y$.
Let $g(u) := x^u - u^x$ for $u \ge x + 1$. We have
$$g'(u) = x^u \ln x - u^{x-1} x
> x^u - u^{x-1}x
= x(x^{u-1} - u^{x-1}).
$$
Let $h(v) := \frac{\ln v}{v - 1}$.
We have $h'(v) = -\frac{v\ln v - v + 1}{v(v-1)^2}$.
Thus, $h(v)$
is strictly decreasing on $v \ge 3$. Thus, $\frac{\ln x}{x - 1}
> \frac{\ln u}{u - 1}$
which results in $x^{u - 1} > u^{x-1}$. Thus, $g'(u) > 0$
for all $u \ge x + 1$.
Thus, we have
$$x^y - y^x = g(y) \ge g(x+1) = x^{x+1} - (x + 1)^x.$$
Let $F(x) := x^{x+1} - (x + 1)^x$.
Using $(1 + 1/x)^x \le \mathrm{e} < x$, we have $(x + 1)^x < x^{x + 1}$. We have
\begin{align*}
 F'(x) &= x^{x+1}\left(\ln x + \frac{x+1}{x}\right) - (x + 1)^x \left(\ln (x + 1) + \frac{x}{x + 1}\right)\\
 &> x^{x+1}\left(\ln x + \frac{x+1}{x}\right) - x^{x+1} \left(\ln (x + 1) + \frac{x}{x + 1}\right)\\
 &= x^{x+1}\left(\frac{1}{x} - \ln(1 + 1/x) + \frac{1}{x + 1}\right)\\
 &> 0
\end{align*}
where we have used $\ln (1 + 1/x) \le 1/x$.
Thus, we have $F(x) \ge F(3) = 3^4 - 4^3 = 17$.
Thus, $x^y - y^x \ge 17$.
We are done.
