... beneath a 🌘 Waning Crescent

…symmetry isn’t just a preference for “pretty” shapes.

Contents

• Introduction
• The HM-AM-GM-QM Inequality
  • The two circles
  • The Semicircle
  • The Container
  • The 3D container
• The Sum of Squares Inequality
• Nesbitt’s Inequality
• Conclusion

Introduction

While I was randomly browsing the web, I came across this nice picture:

Roland H. Eddy, Memorial University of Newfoundland, Canada, 1985

And it tickled my imagination a little, just enough to write this short post.

After writing my previous handout article regarding inequalities, I wanted to see if I could find ways to represent inequalities in a geometrical way (you know, classic circles, triangles, squares, cubes, rectangular prisms and the like). So I’ve been digging and improvising, and I’ve come up with some animations to help people get a geometrical intuition of things that are mostly studied in algebra and analysis.

Some of the animations are standard and are taught in the right kind of schools, but others have some originality. For those, I actually developed the ideas using pen, paper, and my own imagination. If somebody else already did that, it’s fine; I am not a fool to claim “real” originality when it comes to basic mathematics. The roads were already very circulated in the last 2000 years.

The HM-AM-GM-QM Inequality

This is the most popular inequality chain we encounter during our school years. To remind you of it, in case you’ve forgotten, the simple version for three numbers $a, b, c > 0$ is:

\[ \underbrace{\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}}_{\text{HM}} \leq \underbrace{\sqrt[3]{abc}}_{\text{GM}} \leq \underbrace{\frac{a+b+c}{3}}_{\text{AM}} \leq \underbrace{\sqrt{\frac{a^2+b^2+c^2}{3}}}_{\text{QM}} \]

Or, in the even simpler two-variable case:

\[ \frac{2}{\frac{1}{a} + \frac{1}{b}} \leq \sqrt{ab} \leq \frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}} \]

To decode the “alphabet soup,” here is what those letters actually stand for:

• HM = Harmonic Mean: Even if it sounds counterintuitive to an untrained eye, this mean appears in the very laws encoded in our universe. For example, if you go from point $A$ to point $B$ with a speed of $v_1$ and come back with a speed of $v_2$, what is your average speed? A bad student would say $v_{\text{avg}}=\frac{v_1+v_2}{2}$, but a good student would know it is actually the harmonic mean: $v_{\text{avg}} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.

• GM = Geometric Mean: The “growth” mean, useful for scaling and compounding. Much like the HM, this one appears in nature and in… simple finance. For example, if you are a stock investor and in the first year your portfolio grows by 100%, but the next year the market crashes by 50%, what was your average growth?

  • An investor bad at math would say: $\frac{100 + (-50)}{2} = 25%$.
  • A good investor would look at the growth factors: in the first year, the factor is $2.0$; in the second, it is $0.5$. Then it would average everything down like this: $ \text{Average Growth Factor} = \sqrt{2.0 \times 0.5} = 1.0$
  • This means your average growth is actually $0%$. You ended up exactly where you started, which, let’s be honest, is already better than most traders.

• AM = Arithmetic Mean: The classic average everyone knows and loves. Okay, maybe love is a strong word for an average math formula.

• QM = Quadratic Mean: Also known as the Root Mean Square (RMS) appears in electrical engineering for example.

  • In Europe, the voltage is labeled as 230V. But this is not the actual average voltage people think. The actual value is determined with the RMS.

Now that things are clearer, let’s look at this inequality chain with a geometric eye. It’s amazing to see how things come to life.

The two circles

The first animation is actually the one I found in the picture.

We are given a large circle with center $O$ and a diameter $a$, meaning the radius is $R = \frac{a}{2}$. Then, there is another smaller circle with center $O’$ touching the first circle’s circumference from the outside. This second circle has a diameter $b$, so its radius is $r = \frac{b}{2}$. If we project the center of the smaller circle $O’$ onto the vertical line passing through $O$, we name that projected point $P$.

A right triangle $OPO’$ is formed with the following lengths:

• The hypotenuse $OO’$ is the sum of the radii: $\frac{a}{2} + \frac{b}{2} = \frac{a+b}{2}$.
• The horizontal leg $OP$ is the difference of the radii: $\frac{a}{2} - \frac{b}{2} = \frac{a-b}{2}$.
• The vertical leg $O’P$.

To compute $O’P$, we just apply Pythagoras’s theorem:

\[ (O'P)^2 = \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = ab \implies \] \[ \implies O'P = \sqrt{ab} \]

$OO’$ is the AM for $a$ and $b$, while $O’P$ is the GM for $a$ and $b$. Notice how the GM (a leg) is always smaller than the AM (the hypotenuse). In the one particular case where the circles are the same size ($a=b$), the leg $OP$ becomes zero, and the GM coincides with the AM. Lovely!

The Semicircle

This is the “classroom” strategy, the visual I was taught in school.

We start with a semicircle with center $O$ and a total diameter of $a + b$. We pick a point $P$ on the circumference and project it down onto the diameter at point $P’$. This forms a right triangle $POP’$ (the letters were conveniently chosen so the visual “pops”).

In this triangle:

• The horizontal leg is $OP’ = \frac{|a - b|}{2}$.
• The hypotenuse $OP$ is the radius, which is half the diameter: $OP = \frac{a+b}{2}$.

We need to compute the vertical segment $PP’$, and for this, we are going to use Pythagoras’s theorem:

\[ PP' = \sqrt{\left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2} = \sqrt{ab} \]

In essence, we’re looking at the same idea as before, just through a different “mechanic.” We take a diameter of length $a + b$ and split it into two segments, $a$ and $b$.

As you move the separator in the animation, it’s easy to see the machinery at work: the blue line (GM=$PP’$) is always trapped inside the circle, so it can never be taller than the red radius (AM=$OP$). They only hit the same height at the very top, when $a = b$.

Now, to complicate things further, let’s add the QM (Quadratic Mean) into the picture. To do this, we will have to:

• Draw a radius $OM$ that is perpendicular to our diameter $a+b$. Since $OM$ is a radius, we know $OM = \frac{a+b}{2}$.
• Connect points $M$ and $P’$ with a new segment.

By looking at the visual, we can see a new right triangle ($MOP’$) is formed, with its two legs being:

• $OM = \frac{a+b}{2}$
• $OP’ = \frac{|a-b|}{2}$

To compute the hypotenuse $MP’$, we simply apply Pythagoras again:

\[ MP' = \sqrt{\left(\frac{a+b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2} = \sqrt{\frac{a^2+2ab+b^2 + a^2-2ab+b^2}{4}} = \] \[ = \sqrt{\frac{2a^2+2b^2}{4}} = \sqrt{\frac{a^2+b^2}{2}} \]

We observe now that $MP’$ plays the role of the QM of $a$ and $b$. Because $MP’$ is the hypotenuse and $OM$ (the radius/AM) is just a leg, the QM will always be bigger than the radius, unless $a=b$. In that specific case, $P’$ moves to the center $O$, the leg $OP’$ vanishes, and we get QM = AM.

And finally, let’s not forget about the HM. This is the most subtle of them all. To make it “appear,” let’s project the point $P’$ onto the segment $OP$. We will call this new projection point $N$.

To compute $PN$, which is the actual HM, we use the properties of the right triangle $OPP’$. Since $PN$ is a segment on the hypotenuse formed by the altitude from the right angle (wait, actually, we use the area or similarity here!), the math works out beautifully:

\[ PN = \frac{PP'^2}{OP} = \frac{(\sqrt{ab})^2}{\frac{a+b}{2}} = \frac{2ab}{a+b} = \frac{2}{\frac{1}{a} + \frac{1}{b}} \]

We’ve done it! We have the whole “alphabet soup” chain packed into one single semicircle:

• $PN$ is the HM (the small segment)
• $PP’$ is the GM (the vertical altitude)
• $OP$ is the AM (the radius)
• $MP’$ is the QM (the big hypotenuse)

It’s easy to see the hierarchy now. Unless $a=b$, the segments will always stay in their lane: $PN < PP’ < OP < MP’$. Everything is connected. Again, lovely!

The Container

This is not a proof of the AM-GM inequality, but rather a beautiful consequence of it.

The idea for this visual came to me because I recently solved the problem “Container With Most Water”. That coding problem isn’t strictly related to the inequality, but the concept of a container holding water rang a bell…

Think of a container $ABCD$ which is a square. The side of the square is $AB = \frac{a+b}{2}$. This means the area of our square is $\text{Area}_{ABCD} = \left(\frac{a+b}{2}\right)^2$.

Now, let’s introduce a second container: a rectangle $A’B’C’D’$ where the width $A’B’ = a$ and the height $A’D’ = b$. We consider the sum $a+b$ to be fixed, but we start morphing the rectangular container, meaning when we take something from $b$, we put it back into $a$, and vice-versa. The area of this rectangular container is $\text{Area}_{A’B’C’D’} = ab$.

If we want to imagine filling this container with water (I know it’s a 2D shape, get over it), the exact quantity of water that $A’B’C’D’$ can hold is almost always less than what $ABCD$ can hold. No matter how we morph the rectangle, the square will always be the superior vessel.

Look at the following visual (the blue area represents the “liquid” (the product of $a \cdot b$)):

You’ll notice the water level fluctuates as the rectangle morphs (conditioned by $a+b=\text{constant}$), but it’s physically impossible for it to overflow the bounds of the square. The liquid only hits the brim at the “top of the stroke,” right when $a=b$ and the rectangle becomes a perfect square. Any other time, there’s always some empty space left at the top.

We can summarize this “container” experiment with a simple observation:

\[ \text{Area}_{ABCD} \geq \text{Area}_{A'B'C'D'} \implies \left(\frac{a+b}{2}\right)^2 \geq ab \implies \] \[ \implies \frac{a+b}{2} \geq \sqrt{ab} \]

Which is exactly the AM-GM inequality. It’s a physical law of geometry: for a fixed perimeter, the square is the ultimate “water holder,” and the more we stretch the rectangle into a thin line, the more capacity we lose.

The 3D container

For the people not comfortable using a 2D shape as a container, here is the 3D version. The logic still holds, but now, instead of comparing areas, we are going to look at volumes.

We define a cube with a side length of $\frac{a+b+c}{3}$, giving it a volume of $V_{\text{cube}} = \left(\frac{a+b+c}{3}\right)^3$. In parallel, we have a rectangular prism with sides $a$, $b$, and $c$. No matter how we morph that rectangular prism, as long as we keep the sum $a+b+c$ constant, our theoretical blue liquid from the prism will never overflow if we try to pour it into the cube.

Just like before, the “prism” container only reaches the brim of the “cube” container when $a=b=c$. Any deviation from perfect symmetry results in a loss of volume. In mathematical terms:

\[ \left(\frac{a+b+c}{3}\right)^3 \geq abc \implies \frac{a+b+c}{3} \geq \sqrt[3]{abc} \]

What we are seeing here isn’t just a weirdness of algebra, but a fundamental “law of laziness” intrinsic to both the animals and the universe itself. Nature is obsessed with efficiency, and as it turns out, symmetry is the ultimate form of laziness.

The Sum of Squares Inequality

There is a “cute” inequality you might know it simply as the “Sum of Squares”:

\[ a^2 + b^2 + c^2 \geq ab + bc + ca \]

There are multiple ways to prove this, and I’ve already spoken about it in previous posts: here, here, here… and here.

But instead of algebra, we can visualize this as a configuration of squares:

• Imagine a square $ABCD$ with side length $a$;
• Another square $BEFG$ with side length $b$;
• And a third square $EHIJ$ with side length $c$.

The total area of the combined polygon formed by these three is exactly $a^2 + b^2 + c^2$ (the left-hand side of our inequality).

If we draw some new lines, as shown in the animation, new rectangles appear:

• $ABKL$ with area $ca$;
• $ABGM$ with area $ab$;
• $BEJK$ with area $bc$.

As you can see, the sum of these areas is $ab + bc + ca$ and it matches the RHS of our inequality.

As we adjust the sliders to increase $b$ and $c$ to match the size of $a$, the gap between the two sides disappears. The inequality finally becomes an equality when $a = b = c$, and the three squares perfectly align… which is nice I suppose.

Nesbitt’s Inequality

Nesbitt’s inequality is a classical result in the world of mathematics, and it states:

\[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2} \]

If you are curious about its various proofs, I recommend reading about them here, here, and here.

As a challenge, I wanted to see if I could create a visual representation of the Nesbitt structure using basic geometric forms. I “failed” a little, because this specific algebraic structure isn’t exactly geometry-friendly, but with a twist of imagination and some documentation, I’ve come up with this:

The main idea was to play with Viviani’s Theorem. I started by drawing an equilateral triangle, knowing that the sum of the distances from an interior point $Q$ to the sides is always constant.

So, let’s define our segments:

• If we project $Q$ onto $BC$, the length is $x$.
• If we project $Q$ onto $AC$, the length is $y$.
• If we project $Q$ onto $AB$, the length is $z$.

We know that $x + y + z = h$, where $h$ is the altitude of the triangle. To link this to Nesbitt, we perform a variable swap. We let the original Nesbitt variables $a, b, c$ be defined as the sum of these distances: $a = y + z$, $b = x + z$, and $c = x + y$.

When we plug these into the original inequality, something beautiful happens.

Since $y+z$ is just the “total height minus $x$,” and the denominator $b+c$ becomes $(x+z) + (x+y) = (x+y+z) + x = h+x$, our fractions transform into a function of the distance from the sides:

\[ \frac{h-x}{h+x} + \frac{h-y}{h+y} + \frac{h-z}{h+z} \geq \frac{3}{2} \]

In the animation, as you drag point $Q$ around the triangle, you are essentially changing the “balance” of $x, y,$ and $z$. When $Q$ sits perfectly at the center, $x = y = z = h/3$. Each term becomes:

\[ \frac{h - \frac{h}{3}}{h + \frac{h}{3}} = \frac{\frac{2}{3}h}{\frac{4}{3}h} = \frac{1}{2} \]

If $Q$ moves, the symmetry breaks and the sum starts to grow. It’s the perfect visual proof that in the world of Nesbitt, the center is the only place to find the minimum. Lovely!

Even though I said I “failed” to make it perfectly geometry-friendly, this visualization actually reveals something deeper, I cannot put a finger on.

Conclusion

As I tried to force these algebraic structures into circles, triangles, and prisms, I realized something: most algebraic inequalities are not inherently “geometry-friendly.”

Once you move past the basics, the clean lines of basic geometry start to blur. Without the machinery of calculus, derivatives, or multi-dimensional curves, representing a complex algebraic “truth” with just a compass and a ruler feels impossible.

But that is exactly where the beauty lies. By forcing these abstract formulas into a “container” or a “semicircle,” we catch a glimpse of the physical skeleton of mathematics. We see that symmetry isn’t just a preference for “pretty” shapes.