AM-GM-HM inequalities — proofs: Visual walkthrough

Step 0 — What is a "mean", as a picture?

Before any inequality, we need to agree what the three words even mean, using nothing but lengths on a line.

$a$ ::: the first positive number, drawn as a short horizontal stick
$b$ ::: the second positive number, drawn as a long horizontal stick
A "mean" ::: one new stick whose length is decided by a fairness rule applied to $a$ and $b$

We will use exactly $a = 1$ , $b = 4$ for the core pictures (Steps 1–5), so you can compare the three means directly. The edge-case Steps 6 and 7 change the numbers on purpose — and announce it each time.

Step 1 — Build the Arithmetic Mean (the "add-average")

WHAT. Lay stick $a$ and stick $b$ end to end, then fold the total exactly in half.

WHY. The arithmetic mean answers: "what single length, used twice, gives the same total as $a + b$ ?" If you use it twice you must recover the whole; so it is half of the sum. That halving is the folding.

PICTURE. The total length $a + b$ is the whole bar; the midpoint marks where the arithmetic mean sits.

For $a = 1, b = 4$ : $AM = \frac{1 + 4}{2} = 2.5$ . This idea comes straight from an arithmetic progression — equal additive steps.

Step 2 — Build the Geometric Mean (the "multiply-average") as a real height

WHAT. Draw a straight line, put stick $a$ and stick $b$ next to each other on it (total $a + b$ ). Draw a semicircle whose diameter is that whole bar. At the point where $a$ meets $b$ , stand a vertical line up to the circle. Call that vertical height $h$ — its length is the geometric mean.

WHY this picture, and why not just algebra? The geometric mean answers "what single length, used twice by multiplying, gives the same product as $a \cdot b$ ?" — so $GM = ab$ . We need a picture that turns multiplication of two lengths into one length, and the semicircle does it. Here is the reasoning in full so nothing is taken on faith:

Intuition Why the vertical height

$h$ satisfies $h^{2} = a \cdot b$ (proved, not assumed) Two classical facts do all the work. (1) Thales' theorem. Any triangle whose longest side is a diameter of a circle, with its third vertex on the circle, has a right angle at that third vertex. So the triangle from the two ends of the diameter up to the top of our vertical line is right-angled at the top. (This is the standard "angle in a semicircle is $9 0^{\circ}$ " theorem.) (2) The right-triangle altitude relation. In a right triangle, drop the altitude from the right angle onto the hypotenuse. It splits the hypotenuse into two pieces; here those pieces are exactly $a$ and $b$ . The altitude $h$ is the geometric mean of the two pieces: $h^{2} = a \cdot b$ . This follows because the two small triangles are each similar to the whole (they share angles), so $\frac{a}{h} = \frac{h}{b}$ , and cross-multiplying gives $h^{2} = ab$ . Therefore $h = ab = GM$ — the height is literally the geometric mean.

PICTURE. Two coloured segments $a$ and $b$ on the diameter; a vertical violet line of height $h$ rising to the arc; its length is $ab$ .

For $a = 1, b = 4$ : $GM = 1 \cdot 4 = 2$ . This is the geometric progression idea — equal multiplicative steps.

Step 3 — The key comparison: the height can never beat the radius

WHAT. Notice something about the last picture: the semicircle has a radius, and the radius equals the arithmetic mean. Now compare the vertical GM-height $h$ to that radius.

WHY the radius equals AM. The diameter is the whole bar $a + b$ . A radius is half a diameter, so the radius is $\frac{a + b}{2} = AM$ . So the same picture holds both means at once!

PICTURE. Draw the radius from the centre to the point on the arc directly above where the vertical GM stands. That slanted radius (length $AM$ ) is the hypotenuse of the right triangle whose vertical leg is $GM = h$ . A leg of a right triangle can never be longer than the hypotenuse — so $GM \leq AM$ , instantly and visually.

Step 4 — Build the Harmonic Mean inside the same circle

WHAT. From the top of the GM height $h$ , drop a perpendicular onto the slanted radius (the AM-line). The piece of the radius cut off from the centre to that foot is the harmonic mean.

WHY. The harmonic mean is $HM = \frac{2 ab}{a + b}$ . To read it off the geometry we use one more standard fact:

So HM appears as a shorter piece of the AM-line — always shorter than the full GM leg because a projection of a segment can never be longer than the segment itself.

PICTURE. A magenta segment along the radius, clearly shorter than the violet GM leg.

For $a = 1, b = 4$ : $HM = \frac{2 \cdot 1 \cdot 4}{5} = 1.6$ . This is the harmonic progression idea — averaging reciprocals.

Step 5 — All three at once, and the chain reads itself off

WHAT. Stack all three lengths on the same picture: the radius (AM), the vertical GM, and the projected HM.

WHY. Now $GM^{2} = AM \cdot HM$ from Step 4 tells us GM is the geometric mean of AM and HM — so it must sit between them. Combined with Step 3 (GM ≤ AM), the order is forced: $HM \leq GM \leq AM .$

PICTURE. Three horizontal bars of lengths $1.6$ , $2.0$ , $2.5$ — HM shortest, AM longest, GM squarely in the middle.

Step 6 — The degenerate case: when $a = b$

WHAT. Now we deliberately leave the running $a = 1, b = 4$ and slide $b$ down until it equals $a$ . Set $a = b = 3$ and rebuild every mean.

WHY it matters. Every inequality above became an equality somewhere ("leg equals hypotenuse", "projection equals full leg"). We must show the whole machine agrees: all three means coincide.

PICTURE. The meeting point $a ∣ b$ slides to the exact centre. The GM leg becomes the full radius; the HM projection becomes the full leg. All three bars are identical.

Equality in $AM \geq GM \geq HM$ happens if and only if all the numbers are equal.

Step 7 — The other edge: one number shrinks toward zero

WHAT. Again we leave the running values on purpose: keep $a = 4$ fixed but let $b \to 0^{+}$ (a stick shrinking to a dot). Watch all three means.

WHY. The parent note warns AM–GM needs positive inputs. The limit $b \to 0$ shows what "positive but tiny" does, and why the means spread apart maximally.

PICTURE. As $b$ shrinks: AM heads to $a /2 = 2$ , but GM $= 4 b \to 0$ and HM $= \frac{8 b}{4 + b} \to 0$ even faster. The bars fan out — maximum spread, maximum gap.

The one-picture summary

Everything on one diagram: the semicircle on diameter $a + b$ , the vertical GM height, the radius AM (hypotenuse), and the projection HM onto that radius. The single chain $HM \leq GM \leq AM$ is nothing but projection ≤ leg ≤ hypotenuse.

Recall Feynman retelling — explain the walkthrough to a friend

"Take two sticks, a short one and a long one, and lay them in a row. That whole row, folded in half, is the arithmetic mean — the simple 'split it evenly' average. Now bend that row into the flat bottom of a half-circle (the row is the diameter). Stand a pole up from the joint between the two sticks until it touches the curve — that pole's height is the geometric mean, the 'multiply' average. Why? The triangle from the two ends up to the top has a right angle at the top (angle in a semicircle), and the pole is the altitude of that right triangle, which always equals the square root of the two pieces multiplied. Here's the punchline you can see: the pole is a leg of that right triangle whose slanted side is the radius, and the radius is the arithmetic mean. A leg can never be longer than the slanted side, so GM ≤ AM — no algebra needed. Finally, tip the pole over onto that slanted radius; the shadow it casts is shorter still — that shadow is the harmonic mean, which always loses because small numbers dominate it. Make both sticks the same length and everything lines up at the centre: all three means become the same number. Shrink one stick toward nothing and the three spread as far apart as possible, with the harmonic one collapsing fastest. That whole story — projection ≤ leg ≤ hypotenuse — is the inequality AM ≥ GM ≥ HM."

Recall Where this leads next

The same "leg ≤ hypotenuse" idea generalises to power means and to Jensen's inequality.
The product-to-length trick is a cousin of the Cauchy–Schwarz inequality.
Using GM as a fixed floor is the heart of optimisation using inequalities.