Geometric means and the AM-GM inequality

The geometric mean

For two positive numbers $a$ and $b$, the number $G$ that makes $a$, $G$, $b$ a GP is their geometric mean. From $G^2 = ab$,

For $4$ and $9$, $G = \sqrt{36} = 6$, and $4, 6, 9$ is a GP with ratio $\tfrac{3}{2}$. Their arithmetic mean is $\dfrac{4 + 9}{2} = 6.5$, a little more than $6$.

Inserting several means

To insert $n$ geometric means between $a$ and $b$ means to fill in $n$ numbers so that the whole list is a GP. The list has $n + 2$ terms, so $b$ is term $n + 2$:

Example. Insert $3$ geometric means between $5$ and $405$.

Step 1. The ratio. $r^4 = \dfrac{405}{5} = 81$, so $r = 3$ (taking the positive means).

Step 2. The means. $5 \times 3 = 15$, then $45$, then $135$.

Check. $135 \times 3 = 405$, as it should be.

The AM-GM inequality

For any two positive numbers,

with equality only when $a = b$.

Proof. A square is never negative, so $(\sqrt{a} - \sqrt{b})^2 \ge 0$. Expand it: $a - 2\sqrt{ab} + b \ge 0$. So $a + b \ge 2\sqrt{ab}$, and halving gives $A \ge G$. The square is $0$ only when $\sqrt{a} = \sqrt{b}$, that is, when $a = b$.

For $4$ and $16$: $A = 10$ and $G = 8$. For $7$ and $7$: $A = G = 7$.

The picture shows it without algebra. The diameter has length $a + b$, so the radius is $A$. The red half-chord has length $\sqrt{ab} = G$. A half-chord can never be longer than the radius.

Set both sliders to the same number. The two bars become exactly equal.

Two numbers from their means

The arithmetic mean of two numbers is $13$ and their geometric mean is $12$. Find them.

Step 1. Sum and product. $a + b = 2A = 26$ and $ab = G^2 = 144$.

Step 2. A quadratic. The numbers are the roots of $x^2 - 26x + 144 = 0$.

Step 3. Solve. $(x - 18)(x - 8) = 0$, so the numbers are $18$ and $8$.

In general, the numbers are the roots of $x^2 - 2Ax + G^2 = 0$, that is, $A \pm \sqrt{A^2 - G^2}$.

The ratio form. If $A : G = 5 : 4$, write $A = 5k$ and $G = 4k$. The numbers are $5k \pm \sqrt{25k^2 - 16k^2} = 5k \pm 3k$, that is, $8k$ and $2k$. So they are in the ratio $4 : 1$.

Least and greatest values

AM-GM gives a quick lower bound for a sum whose product is fixed.

Example 1. For $x > 0$, $x + \dfrac{1}{x} \ge 2\sqrt{x \cdot \dfrac{1}{x}} = 2$. The least value $2$ comes at $x = 1$.

Example 2. For $x > 0$, find the least value of $4x + \dfrac{9}{x}$.

The product of the two parts is $36$. So $4x + \dfrac{9}{x} \ge 2\sqrt{36} = 12$. Equality needs $4x = \dfrac{9}{x}$, that is, $x = \tfrac{3}{2}$. The least value is $12$.

Example 3. A rectangle has an area of $100$ m². Its perimeter is $2(l + w)$, and $l + w \ge 2\sqrt{lw} = 20$ m. So the perimeter is at least $40$ m, when the rectangle is a square of side $10$ m.

Practise with new numbers

Try it in Python

The program finds both means for several pairs, shows that $A - G$ is never negative, and inserts three geometric means between $5$ and $405$.

Common mistakes

• Using AM-GM with a negative number. It holds only for positive numbers.
• Forgetting when equality holds. A least value counts only if the two parts can be equal.
• Counting the terms wrongly. $n$ means between $a$ and $b$ make $n + 2$ terms, so $b = ar^{n + 1}$.
• Mixing up $G^2$ and $G$. The product of the two numbers is $G^2$, and their sum is $2A$.

Remember these

• $G = \sqrt{ab}$, and $a$, $G$, $b$ is a GP.
• $n$ geometric means between $a$ and $b$: $r = \left(\dfrac{b}{a}\right)^{\frac{1}{n + 1}}$.
• For positive $a$ and $b$, $A \ge G$, with equality only when $a = b$.
• Two numbers with means $A$ and $G$ are the roots of $x^2 - 2Ax + G^2 = 0$.
• A sum of two positive parts with a fixed product is least when the parts are equal.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the geometric mean of $3$ and $27$.

2. Insert $2$ geometric means between $4$ and $32$.

3. Which is larger for $2$ and $8$: the arithmetic mean or the geometric mean? By how much?

4. For $x > 0$, what is the least value of $x + \dfrac{16}{x}$?

5. Two numbers have sum $20$ and geometric mean $8$. Find them.