The Arithmetic Mean–Geometric Mean Inequality

Two means of two numbers

Take two numbers a and b that are not negative. Their arithmetic mean is $\frac{a+b}{2}$(a + b)/2, the usual average. Their geometric mean is $\sqrt{ab}$√(ab), the square root of their product: the side of the square with the same area as a rectangle a by b.

Take a = 2 and b = 8. The arithmetic mean is $\frac{2+8}{2}=5$(2 + 8)/2 = 5, and the geometric mean is $\sqrt{16}=4$√16 = 4. The arithmetic mean is the larger.

Other pairs agree. 9 and 1 give 5 and 3; 49 and 1 give 25 and 7; 3 and 12 give 7.5 and 6. The AM–GM inequality says it always happens: $\frac{a+b}{2}\ge \sqrt{ab}$(a + b)/2 ≥ √(ab) whenever $a\ge 0$a ≥ 0 and $b\ge 0$b ≥ 0, with equality only when a = b. The pairs do not prove that; the proof below does.

Five pairs and their two means. The arithmetic mean is the larger in every row but the last, where the two numbers are equal and so are the means.

The proof is one square

The square of any real number is at least 0. Since a and b are not negative, $\sqrt{a}$√a and $\sqrt{b}$√b exist, and so does their difference. Its square is at least 0: ${\left(\sqrt{a}-\sqrt{b}\right)}^2\ge 0$(√a − √b)² ≥ 0.

Expand the square: ${\left(\sqrt{a}\right)}^2-2\times \sqrt{a}\times \sqrt{b}+{\left(\sqrt{b}\right)}^2=a-2\sqrt{ab}+b$(√a)² − 2 × √a × √b + (√b)² = a − 2√(ab) + b, using $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$√a × √b = √(ab). So $a-2\sqrt{ab}+b\ge 0$a − 2√(ab) + b ≥ 0.

Add $2\sqrt{ab}$2√(ab) to both sides, which moves the middle term across: $a+b\ge 2\sqrt{ab}$a + b ≥ 2√(ab).

Halve both sides

Halving both sides of $a+b\ge 2\sqrt{ab}$a + b ≥ 2√(ab) keeps the inequality the same way round, because 2 is positive. The result is $\frac{a+b}{2}\ge \sqrt{ab}$(a + b)/2 ≥ √(ab): the arithmetic mean is at least the geometric mean.

With 2 and 8: a + b = 10 and $2\sqrt{ab}=2\sqrt{16}=8$2√(ab) = 2√16 = 8, and $10\ge 8$10 ≥ 8. Halved, that is $5\ge 4$5 ≥ 4.

When the two are equal

Each step of the proof can be read in both directions: expanding a square, adding the same amount to both sides, and halving both sides change nothing about which side is larger or whether they are equal. So the two means are equal exactly when the square at the start is 0, ${\left(\sqrt{a}-\sqrt{b}\right)}^2=0$(√a − √b)² = 0.

A square is 0 only when the number squared is 0, so $\sqrt{a}=\sqrt{b}$√a = √b, and that means a = b. When a and b differ, the square is positive, and the arithmetic mean is strictly the larger.

One number 0 and the other positive is a case where they differ: with a = 0 and b = 6, the arithmetic mean is 3 and the geometric mean is 0.

A semicircle on a diameter of length a + b = 8, split into a and b at the handle. The radius, 4, is the arithmetic mean. The height of the semicircle above the split is $\sqrt{ab}$√(ab), the geometric mean: at a = 2 and b = 6 it is $\sqrt{12}\approx 3.464$√12 ≈ 3.464. No height of a semicircle is more than its radius, and the two are equal only when the split is at the center, a = b = 4.

Why the height is the geometric mean

Join the top of the height to the two ends of the diameter. The angle at the top is a right angle, because it stands in a semicircle. The height then splits that right triangle into two smaller triangles with the same angles, so their sides are in proportion: $h/a=b/h$h / a = b / h. Multiplying out, $h^2=ab$h² = ab, so $h=\sqrt{ab}$h = √(ab).

The radius is half the diameter, $\frac{a+b}{2}$(a + b)/2. The height is half a chord of the circle, and no chord is longer than a diameter, so $h\le \frac{a+b}{2}$h ≤ (a + b)/2: the same inequality, seen in a circle.

Finding a least value

Equality is how the inequality finds a least value. Read $a+b\ge 2\sqrt{ab}$a + b ≥ 2√(ab) in words: when two positive numbers have a fixed product, their sum is at least twice the square root of that product, and the sum is smallest exactly when the two numbers are equal.

Two positive numbers multiply to 36. Their sum is at least $2\sqrt{36}=12$2√36 = 12, and 6 and 6 make it exactly 12, so 12 is the least sum. 4 and 9 also multiply to 36, and their sum is 13.

For x > 0, the numbers x and $\frac{1}{x}$1/x multiply to 1, so $x+\frac{1}{x}\ge 2\sqrt{1}=2$x + 1/x ≥ 2√1 = 2, with equality only when $x=\frac{1}{x}$x = 1/x, that is, at x = 1. At x = 4 the sum is $4+\frac{1}{4}=4.25$4 + 1/4 = 4.25, and at $x=\frac{1}{2}$x = 1/2 it is 2.5.

A rectangle of area 36 with width w has length $\frac{36}{w}$36/w and perimeter $2\left(w+\frac{36}{w}\right)$2(w + 36/w). Since $w+\frac{36}{w}\ge 2\sqrt{36}=12$w + 36/w ≥ 2√36 = 12, the perimeter is at least 24, and it is 24 only when $w=\frac{36}{w}$w = 36/w, which is the 6 by 6 square.

Rectangles of area 36. The perimeter falls as the rectangle gets closer to a square, and is least, 24, for the 6 by 6 square.

The usual mistakes

Mixing up the two means. $\frac{a+b}{2}$(a + b)/2 is the average; $\sqrt{ab}$√(ab) is the square root of the product.

Using it with negative numbers. The proof needs $\sqrt{a}$√a and $\sqrt{b}$√b, so a and b must not be negative. For a = −2 and b = −8 the average is −5, while $\sqrt{ab}=\sqrt{16}=4$√(ab) = √16 = 4 is larger.

Expecting equality when one number is 0. With a = 0 and b = 6 the means are 3 and 0.

Taking the bound as the least value when the two terms cannot be made equal. The bound $2\sqrt{ab}$2√(ab) is reached only when the two terms are equal; if a condition of the problem rules that out, the least value is above the bound.

A pen and a round trip

In the applications below, a pen against a barn wall needs fencing on three sides, and the AM–GM inequality finds the least fencing and then the cheapest fence, each where its two terms are equal. Then a cyclist’s average speed out against the wind and back is shown never to exceed the mean of her two speeds.