Can someone help me finding a lower bound to the function
$$f(x)=\frac{x}{x^{1/x}},$$
where $x\in[3,+\infty[$?
I suppose that a lower bound function can be $y=x$ but I don't really know how to start, so any help would be most welcome.
calculusfunctionsoptimization
Can someone help me finding a lower bound to the function
$$f(x)=\frac{x}{x^{1/x}},$$
where $x\in[3,+\infty[$?
I suppose that a lower bound function can be $y=x$ but I don't really know how to start, so any help would be most welcome.
Best Answer
As indicated in my comment, you can show that $g(x)=x-\log x$ is a lower-bound function to $f(x)=x/x^{1/x}$.
For a proof, observe that we can write $g(x)=x\left(1-\log x/x\right)$ and $f(x)=x\exp\left(-\log x/x\right)$. Since $\exp y>1+y$ for all real $y$ it follows, by setting $y=-\log x/x$, that $\exp\left(-\log x/x\right)$ is greater than $1-\log x/x$ and thus $f(x)>g(x)$ for $x>0$.
In fact, we can also show that the difference between $f(x)$ and $g(x)$ approaches zero as $x$ tends to infinity. This follows from the expansion $\exp y=1+y+\frac12y^2+\mathcal{O}_{y\to 0}(y^3)$ which implies that $$ \exp\left(-\log x/x\right) - \left(1-\log x/x\right) = \frac12\left(\frac{\log(x)}{x}\right)^2 + \mathcal{O}_{x\to\infty}\left(\left(\frac{\log(x)}{x}\right)^3 \right). $$ In particular, $f(x)-g(x)\sim_{x\to\infty}\frac12\log x^2/x$.
Numerically, the maximum of $f(x)-g(x)$ for $x>1$ occurs at approximately $x=8.1$ with a value of approximately $0.25$, so the approximation is indeed quite good.