Resolve $x+\frac{1}{x}$ While reading an old mathematics question paper of the Unified International Mathematics Olympiad, I hit a question with two very interesting patterns. Here I first mention them before presenting my question.
Pattern 1

When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$.

Here, the value of $x$ and $x+\frac{1}{x}$ are in inverse proportion. I tried putting the values and here are the results:
$$0.000001 + \frac{1}{0.000001} = 1000001$$
$$0.999999 + \frac{1}{0.999999} = 2.000001000001$$
The value of $x+\frac{1}{x}$ got near about two, but not equal to it.
Pattern 2

When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$.

This pattern also works the same way as the first one did. Its just that here the value of $x+\frac{1}{x}$ will always evaluate to a number small than $2.5$.
$$1.000001 + \frac{1}{1.000001} = 1.999999000001$$
$$1.999999999999999 + \frac{1}{1.999999999999999} = 2.499999999999999$$
The value of $x+\frac{1}{x}$ got near about $2.5$, but not equal to it.
Question
How can we prove the statements, "When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$" and "When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$." mathematically?
Any help is appreciated. Thank you!
 A: $x+\frac{1}{x}\geq2$ holds for all $x>0$, with equality at $x=1$ only. As a hint, expand
$$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq0.$$
For the second claim, observe that for $1<x<2$, we have
$$(x-2)(x-1)<0\implies x^2+2<3x\implies x+\frac{1}{x}<\frac{x}{2}+\frac{3}{2}<\frac{5}{2}.$$
A: I just have a problem with one thing you wrote. You have

$$0.000001 + \frac{1}{0.000001} = 1000001$$

But without doing any calculations, I know this can't be true. We have here
$$10^{-6} + \frac{1}{10^{-6}} = 1000001$$
$$10^{-6}+10^6=1000001$$
But that means
$$\text{(not an integer)}+\text{(integer)}=\text{(integer)}$$
Which is impossible.
A: If you're familiar with calculus, there's a very snappy way to analyze this sort of situation: taking derivatives (and thinking about what the derivative says about the function in question).
The derivative of $f(x)=x+{1\over x}$ is $1-{1\over x^2}$. For $x\in (0,1)$ this is negative, so $f$ is decreasing on $(0,1)$. In particular, this means that for each $a\in (0,1)$ we have $f(a)>f(1)$. Now just calculate $f(1)$.
This also works for analyzing the interval $(1,2)$. In this interval - and indeed for all $x>1$ - the derivative of $f$ is positive and so $f$ is increasing. So e.g. for every $x\in (1,2)$ we have $$f(1)<f(x)<f(2)$$ or more snappily $$1<f(x)<2.5.$$
Similarly, we get $$f(17)<f(x)<f(42),$$ or $$f(x)\in (17+{1\over 17}, 42+{1\over 42}),$$ for all $x\in (17,42)$. And so on.
A: $
x+\dfrac{1}{x}
$ is increasing on $[1,2]$. So the maximum is attained when $x=2$.
