TITLE: Visualize $z+\frac{1}{z} \ge 2$
QUESTION [5 upvotes]: As we know, always
$$z+\frac{1}{z} \ge 2,~~~~~~~~~ z\in \mathbb{R}^+$$
However, is there any geometric way to visualize this equation for some one who is not that expert in math?
I know this question might have various answers.

REPLY [3 votes]: Plot $xy = 1$ and $x+y = 2$.

Interpret $x=z$ and $y=1/z$ (which can only hold on the blue hyperbola). Or vice versa. The fact the plot is symmetric under swapping $x$ and $y$ is important! (i.e. symmetric under reflecting across the line $x=y$)
It may help to see some more lines depicting how $x+y$ varies:

The red line $x+y=2$ is the smallest line that still meets the hyperbola, and the symmetry makes it easy to find the point of intersection: it's the point where $z = 1/z$.
The above assumes you mean to consider only positive $z$. If you allow negative $z$, then you get the same picture but flipped in the third quadrant. (and it corresponds to $z + 1/z \leq -2$)

Rotating the figure might help, so that $x+y$ is along the vertical axis. If we use coordinates $(u,v) = (x-y, x+y)$:

On the hyperbola, we have
$$ z = \frac{v \pm u}{2} \qquad \qquad \frac{1}{z} = \frac{v \mp u}{2} $$
(either choice of sign is fine; remember the symmetry!)
That is, the hyperbola is the equation
$$ \frac{v \pm u}{2} \cdot \frac{v \mp u}{2} = 1 $$
or more simply,
$$ v^2 = 4 + u^2 $$
Now it's even algebraically obvious what range of values $v$ can take, since $u^2$ can be any (and can only be a) nonnegative number!