some basics

Well, this problem can be solved using calculus to find the points at which the function $f(x)=x+\dfrac{1}{x}$ has $f'(x)=0$ and then use the second derivative test to find the points of maxima and minima. But since this problem is in the Algebra section, I will illustrate a way using algebra to get the answer.

$f(x)=x+\dfrac{1}{x}=(\sqrt{x}-\dfrac{1}{\sqrt{x}})^2+2$

Now, $(\sqrt{x}-\dfrac{1}{\sqrt{x}})^2\geq 0 \quad \forall x\in \mathbb{R^+}$ . The minimum value will be obtained if $(\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=0$ which, if you observe, is attained at $x=1\in \mathbb{R^+}$ . So, minimum value of $f(x)$ when we are putting the restriction $x\in \mathbb{R^+}$ will be at $x=1$ with $f(1)=\boxed{2}$

$x^2 + 1 >= 2x \Rightarrow x + \frac{1}{x} >=2$