Show that if $x \ge 1$, then $x+\frac{1}{x}\ge 2$ So here the problem goes:

Show that if $x \ge 1$, then $x+\frac{1}{x}\ge 2$.

This is a very interesting word problem that I came across in an old textbook of mine. So I know it's got something to do with 
If $x = 1,$ then we have $1 + \frac 11 = 2,$
and clearly, if $x \ge 2,$ then since $\frac 1x$ is also positive, $x+\frac{1}{x}\ge 2$. 
So we only have left $1 < x < 2$ to deal with.
But other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :)

Since $x\gt 0$ and $\frac 1x\gt 0$ for $x\ge 1$, by AM-GM inequality, we have
$$x+\frac 1x\ge 2\sqrt{x\cdot\frac 1x}=2.$$
