The AM-GM Inequality, Proved Geometrically With a Semicircle and √(ab)

The arithmetic mean of two positive numbers a and b is \dfrac{a+b}{2}. The geometric mean is \sqrt{ab}. Both of them are natural "averages" — the first is the average of sums, the second is the average of products. A deep fact, true for every choice of a and b, is that the arithmetic mean is always at least as large as the geometric mean:

\frac{a + b}{2} \geq \sqrt{ab}.

You can prove this algebraically in three lines. But there is a picture that proves it without algebra at all — and the picture is built out of a semicircle, a diameter, and a single perpendicular of length \sqrt{ab}.

The setup

Lay out a straight segment of length a + b on a number line. Call the left end L, the right end R, and let P be the point that splits the segment into a piece of length a (on the left) and a piece of length b (on the right).

Draw a semicircle with diameter LR. Its radius is exactly (a+b)/2 — the arithmetic mean.

Now, at the point P, erect a perpendicular upward until it meets the semicircle at a point Q. The length PQ, as you will see in a moment, is exactly \sqrt{ab} — the geometric mean.

The semicircle of diameter $a+b$ has radius $(a+b)/2$ — the arithmetic mean. The perpendicular from $P$ to the semicircle has length $\sqrt{ab}$ — the geometric mean. The radius is the *longest* perpendicular you can draw from the diameter to the arc, so $\text{AM} \geq \text{GM}$, with equality only when $P$ sits exactly at the centre $O$, which happens precisely when $a = b$.

Why the perpendicular has length √(ab)

This is the step that pulls the geometric mean out of thin air. It comes from a classical theorem: if a right-angled triangle is inscribed in a semicircle with the hypotenuse as the diameter, then the altitude from the right angle to the hypotenuse has length equal to the geometric mean of the two segments the altitude creates.

Concretely: connect L to Q and R to Q. The angle LQR is a right angle, because an inscribed angle that stands on a diameter is always 90° (Thales's theorem).

Now triangles LPQ and QPR are both similar to the larger right triangle LQR. Why: each shares an angle with the larger triangle and also has a right angle at P, giving two equal angles and hence similarity. From the similarity LPQ \sim QPR:

\frac{LP}{QP} = \frac{QP}{PR} \quad\Longrightarrow\quad (QP)^2 = LP \cdot PR = a \cdot b.

So QP = \sqrt{ab}, exactly.

Why: the equal ratio comes from matching corresponding sides in the similar triangles. Cross-multiplying gives (QP)^2 = a \cdot b, so the length QP is the positive square root of ab.

Why AM ≥ GM, on the picture

The segment OQ from the centre O to the top of the semicircle is a radius, so it has length (a+b)/2 — the arithmetic mean. The segment PQ is a perpendicular from a different point on the diameter.

A simple fact about circles: the longest perpendicular you can drop from any point on the diameter to the arc is the one at the centre, because the arc is highest directly above the centre. Every other perpendicular is shorter.

In a right triangle with hypotenuse OQ and one leg PQ, the leg is always at most the hypotenuse:

\sqrt{ab} = PQ \leq OQ = \frac{a+b}{2}.

Equality holds only when P = O, which is precisely when the two pieces are equal — that is, a = b. The picture makes the inequality visible and the equality condition obvious.

Test it with numbers

Drag P to try a = 4, b = 1 (so a + b = 5). The arithmetic mean is 2.5 and the geometric mean is \sqrt{4 \cdot 1} = 2. The radius is 2.5; the perpendicular is 2. The radius is longer.

Try a = b = 3. Both means are 3. The perpendicular sits at the top of the semicircle and has length 3 — same as the radius. Equality.

Try a = 9, b = 1. Arithmetic mean is 5; geometric mean is \sqrt{9} = 3. The radius is 5; the perpendicular is 3. The gap is visible.

A two-line algebraic proof for comparison

The picture proves it. So does algebra. Start from the fact that (\sqrt{a} - \sqrt{b})^2 \geq 0 (a square of a real number is non-negative). Expand:

a - 2\sqrt{ab} + b \geq 0 \quad\Longrightarrow\quad a + b \geq 2\sqrt{ab} \quad\Longrightarrow\quad \frac{a+b}{2} \geq \sqrt{ab}.

Why: moving 2\sqrt{ab} to the right side is legal because inequality is preserved under adding the same number to both sides. Dividing by 2 preserves the inequality because 2 > 0.

Equality holds iff \sqrt{a} = \sqrt{b}, iff a = b. Same equality condition as the picture.

Why this inequality shows up everywhere

The AM-GM inequality is one of the most useful tools in JEE algebra problems. Whenever you see a minimum-or-maximum question with a product or a sum constraint, AM-GM is the first technique to reach for. A classic example: if a + b = 10 and a, b \geq 0, what is the largest possible value of ab? The AM-GM bound says ab \leq \left((a+b)/2\right)^2 = 25, attained at a = b = 5.

The inequality generalises beyond two numbers. For n positive reals a_1, a_2, \dots, a_n:

\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}.

A visual proof of the general n-variable version is harder, but the two-variable picture is the kernel. See AM, GM, HM Inequalities for the full story and its proofs.