How to proof uniqueness with the usage of function and proving an inequality My first problem is connected with showing that there is only one couple of natural numbers which fulfills the condition $$x^y=y^x, \ x<y.$$ Of course these numbers are 2 and 4. It can be done with the help of the function $f(x)=x^{\frac{1}{x}}$. Unfortunately I don't see any connection here.
Second problem associated with the first one is with proving the inequality
$$ \left(\dfrac{33}{32}\right)^{32}<\dfrac{32}{3}.$$
I can't prove in explicitly. Intuitively it is quite clear, but I seek formal proof which does not come to my mind.
 A: Hint for the first problem
Note that $x^y = y^x$ can be written as $x^{1/x} = y^{1/y}$. The claim is equivalently asking for $x,y$ with $x < y$ so that $f(x) = f(y)$. You want to show that $(x,y) = (2,4)$ is the only solution.
To do this, try using calculus to show that $f$ does not take on the same value twice (after the exceptions $2,4$). There is a specific function property which is helpful here. It might also help to compute some small values of $f$ (or even better, use software to plot $f$) to get an idea for how the function behaves.
Hint for the second problem
Never mind, just do what the other answerer says.
It looks like they want you to rewrite this in terms of $f$, but honestly, the first solution method that pops into my mind is to use the binomial theorem and bound small terms. In other words, do something like this:
$$ (1 + \frac{1}{32} )^{32} \leq 1 + \frac{1}{32} \cdot 32 + \left( \frac{1}{32} \right)^2 \binom{32}{2} + \dots $$
A: For $x>0$ we have
$$ \left(1+\frac{1}{x}\right)^x<e$$
since the function $f(x)=\left(1+\frac{1}{x}\right)^x$ is increasing on $(0,\infty)$ and approaches $e$.
Therefore
$$ \left(1+\frac{1}{32}\right)^{32}<e<\frac{32}{3} $$
