How to prove this inequality $ x + \frac{1}{x} \geq 2 $ I was asked to prove that: 
$$x + \frac{1}{x}\geqslant 2$$
for all values of $ x > 0 $
I tried substituting random numbers into $x$ and I did get the answer greater than $2$. But I have a feeling that this is an unprofessional way of proving this. So how do I prove this inequality?

Obviously there's absolutely no shortage of methods for doing this; here's yet another.  As in other answers, we can assume that $x\geq 1$ (otherwise, swap $x$ for $\frac1x$); now set $y=x-1$ (and note that we have $y\geq 0$).  It's easy to show that $\frac1{1+y}\geq 1-y$, since multiplying both sides by $1+y$ we get $1\geq (1-y)(1+y)=1-y^2$.  But then this means that $x+\frac1x$ $=1+y+\frac1{1+y}$ $\geq 1+y+(1-y)$ $=2$.
The main takeaway from this proof, to me, isn't the result itself, but the inequality at the heart of it: what might be called the "Geometric Inequality" $\frac1{1+y}\geq 1-y$ for $y\geq 0$ (and similarly, $\frac1{1-y}\geq 1+y$ as long as $y\lt 1$) has broad applicability for inequalities and estimations.
