Visualize $z+\frac{1}{z} \ge 2$ 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. 
 A: The Arithmetic Mean-Geometric Mean Inequality states that for any non-negative numbers $a$ and $b$, we have $AM := \dfrac{a+b}{2} \ge \sqrt{ab} =: GM$. 
This can be visualized as follows: 
Now, set $a = z$ and $b = \dfrac{1}{z}$ to visualize what you wanted. 
A: 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!
A: Well, what you wrote is clearly not true, it is only true for $x\in (0,\infty)$.
You could just try plotting the function $x\mapsto x+\frac1x$ and see that it is always above $2$.
http://www.wolframalpha.com/input/?i=plot+z+%2B+1%2Fz
Or, you can plot $x\mapsto x$ and $x\mapsto \frac1x$ and try to understand what is happening on $(0,\infty)$. It's clear you only need to look at the interval $(\frac12, 2)$, since the equality obviously holds outside it.
https://www.wolframalpha.com/input/?i=plot+x+and+1%2Fx+from+x%3D1%2F2+to+2
Now, on this graph, I think it's possible to explain what is happening:


*

*It is clear that for $x=1$, the sum of the two function is precisely $1$.

*If you go to the right of the point $x=1$, then the bottom function $\frac1x$ is decreasing more slowly than the top function $x$ is increasing, so, their sum will be above $2$, because it will be $$(1 + \text{something}) + (1 - \text{something smaller}) = 2 + (\text{something} - \text{something smaller})$$ so it will be $2$ plus something positive.

*If you go to the left of the point $x=1$, then the bottom function $x$ is decreasing more slowly than the top function $\frac1x$ is increasing, so, their sum will be above $2$, because it will be $$(1 + \text{something}) + (1 - \text{something smaller}) = 2 + (\text{something} - \text{something smaller})$$
so it will be $2$ plus something positive. 

A: 

For plotting y=x+1/x just shift the y=1/x graph above/below y=x line and you're done!
Here are the plots of y=x,y=1/x and y=x+1/x.Adding the first two graphs gives you the third.I guess you can visualize it "graphically" now. BTW i guess @Jimmy has shown the best geometric method out there.
A: In 2D space, consider a rectangular container filled with $1$ unit of incompressible fluid. If its width is $z$, then its height is $1/z$, and its perimeter is $2 + 2/z$. Suppose the walls of the container are elastic, so that it tries to minimize its perimeter, but it is still constrained to take the shape of a rectangle.
Imagine that the rectangle starts with dimensions $0.2 \times 5$. What will it do? It will snap inwards along its longer dimension, bulging outwards along its shorter dimension, until it settles on a $1\times 1$ square. That square attains the minimum perimeter, which is $4$, representing the inequality $2 + 2/z\geq4$.
Can you visualize the rectangle going KSSSschlllurp! and finding that minimum? Hopefully you can, because I'm too lazy to produce the right computer animation. :)
Bonus: Once you have that animation, you can also overlay the plots described in the other answers! In particular, you can show the hyperbola along which a corner of the rectangle travels, and you can show the gradient of the potential $x+y$. This connects the visualization to a more algebraic approach.
A: For $z>0$, the derivative of the function $x(z)=z+1/z$ has the only zero at $1-1/z^2=0$, that  is $z=1$ and the second derivative $2/z^3>0$, so the function is convex. Therefore, it's global minimum is at $z=1$, that is $x(1)=2$ and $x(z)\ge 2$. 
