Prove or disprove this inequality for $xProve or disprove this inequality for $x<y$:
$$(2^{x}-1)^{y}<(2^{y}-1)^{x}$$
I assume that $y=x+r$ for $r>0$ and I consider the function $$f(x)=(2^{x}-1)^{x+r}-(2^{x+r}-1)^{x}$$
 I wante to see the behavior of $f$ with respect to $x$. I've calculated the derivative but it is very complicated to determine its sign. The derivative is: $$ln((((2^{x}-1)^{(2^{x}-1)^{r+x}})/((2^{r+x}-1)^{(2^{r+x}-1)^{x}}))(((2^{2^{x}r(2^{x}-1)^{r+x-1}})(2^{ 2^{x}x(2^{x}-1)^{r+x-1}}))/(2^{2^{r+x}x(2^{r+x}-1)^{x-1}})))$$
 A: The inequality doesn't make sense for negative values of the variables, so we can assume that $0\lt x\lt y$ is intended, in which case the inequality amounts to saying that the function
$$f(x)={1\over x}\ln(2^x-1)$$
is increasing for $x\gt0$.  Can you take it from here?
A: I assume both $x$ and $y$ are positive, to avoid nonreal values.
Analysis. Letting $p=\frac yx>1$, the desired inequality would follow from
$$ (2^x-1)^p \le 2^{px}-1 $$
Equivalently,
$$ p\ln(2^x-1) \le \ln(2^{px}-1) $$
This suggests considering the function $f(x)=\ln(2^x-1)$.  Here is a (qualitatively roughly correct, but quantitatively quite inaccurate) sketch:

Based on this sketch we certainly expect that $pf(x)\le f(px)$.  One way to formulate the relevant properties of the sketch rigorously is to say that $(px,f(px))$ is closer to the asymptote $y=x\ln 2$ than $(x,f(x))$ is, while $(px,pf(x))$ is further away.  This approach yields the following argument.
Proof. Check that
$$ t\ln 2 - \ln(2^t-1) = \ln\left(\frac{2^t}{2^t-1}\right) $$
is a strictly decreasing and positive for $t>0$.  Thus $x<y$ implies
\begin{align*}
y\ln 2 - \ln(2^y-1)
&< x\ln 2 - \ln(2^x-1) \\
&< \frac yx (x\ln 2 - \ln(2^x-1)) \\
&= y\ln 2 - \frac yx \ln(2^x-1)
\end{align*}
Now simplify.
