Is the function $f(x) = x + (1/x)$ always unbounded for $x > 1$? PROBLEM

Is the function $f(x) = x + \frac{1}{x}$ always unbounded for $x > 1$?

MY ATTEMPT
The first derivative is
$$f'(x) = 1 - \frac{1}{x^2}$$
which is positive for $x > 1$.
The second derivative is
$$f''(x) = \frac{2}{x^3}$$
which is positive for $x > 1$.
Thus, the function $f(x) = x + (1/x)$ is increasing and concave up, for $x > 1$.
To see the general trend, I graphed the function using Desmos and got the following image:

QUERY


My problem is that I have trouble seeing whether this result implies that $f$ is unbounded for $x > 1$ in full generality?


Sure, I know that
$$\lim_{x \to \infty}{f(x)} = \infty$$
$$\lim_{x \to 0^+}{f(x)} = \infty,$$
but what about the cases in between?
Thanks!
 A: (I think that the OP's  comment above helped clarify the question.)
A function is bounded or unbounded on some interval (or some set).
When we say that some function is unbounded without specifying the interval (or set), we mean on its domain.
The function you are asking about is unbounded (on its domain).
On any bounded interval with positive lower bound, the function is bounded.
So, on $\left(\frac{3}{29}, 95\right)$ and on $(2,137)$ and on $[6,6.003]$ the function is bounded.
On $(0,5]$ and on $(5,\infty)$ and on $(873,\infty)$ it is unbounded.
A: Your question has the word "always" in it.  That points to a misconception.  If I had a function $f(x)$ for which $\lim_{x \to \infty} f(x) = 3$, I would not ask if $f(x)$ was "always" $3$.    It could be that $f(x)$ is never $3$, but only asymptotic to it.
So when we say $f(x)$ is "unbounded" we mean it approaches infinity somewhere, not everywhere (or "always".)  Your function is unbounded on $1<x<\infty$, but at no point in that interval is the value of $f(x)$ actually equal to infinity.
A: HINT: using $AM-GM$ we get (since $x>1$) we have $$x+\frac{1}{x}\geq 2$$
A: Yes, of course. Try $x\rightarrow+\infty$.
We obtain:
$$x+\frac{1}{x}=x\left(1+\frac{1}{x^2}\right)\rightarrow+\infty.$$
