Although this particular question is most elegant when thought as about arithmetic and geometric means (as seen in some other answers), there is a systematic way to solve inequalities like this that nobody seems to have brought up.
To solve an inequality of the form $f(x) \leq g(x)$ (or with $<$, $\geq$, etc), first turn it into an equation, and find its solutions (hopefully a finite set). Then find every place where $f$ or $g$ is discontinuous (hopefully a finite set), and every place where $f$ or $g$ is undefined (hopefully a finite set or a finite union of intervals, and we actually just need the endpoints of the intervals). You now have a set of $x$ values (hopefully finite, with cardinality $n$), which divide the number line into intervals ($n+1$ of them). Pick a number from each interval. Test whether the inequality is true for each number that you have (all $2n+1$ of them). If you picked the number from an interval, then your test applies to all numbers in that interval; now you have the complete answer.
In pre-Calculus classes, since we don't officially know about discontinuity, I teach this method with a list of allowed operations to appear in the formulas for $f$ and $g$; then I state where these might be discontinuous. Similarly, the idea of considering where $f$ and $g$ are undefined amounts to identifying the specific operations to look at. See http://tobybartels.name/MATH-1150/2018FA/inequalities/ for a fuller description.
In this case, solve $x + 1/x = 2$ by multiplying by $x$ to clear fractions, getting the quadratic equation $x^2 + 1 = 2x$, whose only solution (found by factoring, completing the square, or the quadratic formula) is $x = 1$. In the original inequality, there is a division by $x$, so we also need to solve $x = 0$, which is immediate. Otherwise, everything is defined and continuous. Choose a number less than $0$, such as $-1$ (or skip this since the original question specified $x > 0$); a number between $0$ and $1$, such as $1/2$; and a number greater than $1$, such as $2$. The inequality is false when $x = -1$ (since $-2 < 2$), false (or meaningless) when $x = 0$ (because of division by zero), true when $x = 1/2$ (since $5/2 > 2$), true when $x = 1$ (since $2 = 2$), and true when $x = 2$ (since $5/2 > 2$ again). Therefore, it is true for precisely those $x$ strictly between $0$ and $1$, equal to $1$, or greater than $1$; in other words, for $x > 0$.
Now you've not only proved the statement but also justified the hypothesis that $x > 0$; without that, it would be false. And if you want to know when the inequality is tight, I found that in the very first step (precisely when $x = 1$).