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!

$x+\frac{1}{x}\geq2$ 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
$$(x-2)(x-1)<0\implies x^2+2<3x\implies x+\frac{1}{x}<\frac{x}{2}+\frac{3}{2}<\frac{5}{2}.$$
