Find lower bound of function $\frac{x}{x^{1/x}}$ 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.
 A: 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.

A: Consider the monotonic transformation
$$g(x) = \ln f(x) = \ln x - \frac 1x \ln x = \ln x \cdot \left(1- \frac 1x\right)$$ 
$g(x)$ is obviously the product of two strictly increasing functions of $x$. So its lower value is attained at the minimum value $x$ can take, i.e at $x=3$. Since moreover $g(x)$ is a monotonic transformation of $f(x)$, the same holds for $f(x)$.
So, for the given domain of $x$,
$$\min f(x) = f(3) = \frac {3}{3^{1/3}} = 3^{2/3}$$
ADDENDUM
The OP looks for a lower-bound function rather than the function's minimum. Then we are looking for some 
$$h(x) : f(x) \ge h(x) \Rightarrow \frac{x}{x^{1/x}} \ge h(x) \Rightarrow x-h(x)x^{1/x}\ge 0$$
Try $h(x) = x^a$. Then we must have
$$x-x^ax^{1/x}\ge 0 \Rightarrow x \ge x^{a+\frac 1x} \Rightarrow 1 \ge a +\frac 1x \Rightarrow a\le 1- \frac 1x$$
Since $\min x = 3$ we can set $a=2/3$ and so a lower bound function is
$$h(x) = x^{2/3}$$
But this becomes very loose very quickly.
