# 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.

• That's not true for $z<0$. Maybe you mean $z\in\mathbb R^+$. Anyway, for a geometrical interpretation, one can compare the graph of $y=1/x$ and $x+y=2$: wolframalpha.com/input/?i=plot%20y%3D1%2Fx%2C%20x%2By%3D2
– user856
May 25, 2015 at 5:56
• @Rahul Thanks for mentioning that. May 25, 2015 at 8:15

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.

• IMO this is the best answer so far (the others are pretty much just "plot it") but not perfect. If there was some way to integrate the $z$ and $\frac{1}{z}$ more naturally instead of just saying "set $a = z, b = \frac{1}{z}$ then that would be perfect.
– MCT
May 25, 2015 at 6:03
• Since the geometric mean of z and 1/z is 1, erect a unit line prependicular to the end of a segment of length z. Connect the other end of the segment to the end of the unit segment, and draw the perpendicular to it. That will intersect the extended segment of length z in a segment of length 1/z. May 25, 2015 at 6:22

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!

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.

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.

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.

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 $x(z)\ge 2$.

• But the OP is asking for a way to visualize this statement; this offers no insight on it.
– MCT
May 25, 2015 at 6:02
• Convexity is a geometric concept...
– DVD
May 25, 2015 at 6:04
• Yeah, well so is everything else. If you want to highlight the geometric concept of convexity, then don't make your answer entirely algebraic.
– MCT
May 25, 2015 at 6:04
• Not everything else...
– DVD
May 25, 2015 at 6:06
• You understand my point though hopefully. Many things have a geometric analogue, and if you want to use that analogue as a way to visualize a concept, then your answer ought to actually touch on that instead of just using algebra then essentially saying "it is geometric, trust me" and calling it a day. (-1)
– MCT
May 25, 2015 at 6:14