Geometric means

The arith­metic mean of two num­bers sits halfway between them by addi­tion. The geo­met­ric mean sits halfway between them by mul­ti­pli­ca­tion: you mul­ti­ply by the same fac­tor to go from the first num­ber to the mean as to go from the mean to the sec­ond. This les­son defines the geo­met­ric mean, shows how to insert sev­eral geo­met­ric means between two num­bers, and proves the famous fact that the arith­metic mean is never smaller than the geo­met­ric mean.

Through­out this les­son the num­bers are pos­i­tive.

Geo­met­ric mean of two num­bers#

If $a$, $G$, $b$ are in G.P., then $G$ is called the geo­met­ric mean (G.M.) of $a$ and $b$. By Prop­erty 1 of a G.P., $G^2 = ab$, so

$$G = \sqrt{ab}$$

For exam­ple, the G.M. of 4 and 25 is $\sqrt{100} = 10$, and indeed $4, 10, 25$ is a G.P. with com­mon ratio $\displaystyle \frac{5}{2}$.

Geo­met­ric mean of sev­eral num­bers#

The geo­met­ric mean of $n$ pos­i­tive num­bers is the $n$th root of their prod­uct:

$$G = \sqrt[n]{a_1 a_2 \cdots a_n}$$

For exam­ple, the G.M. of 2, 4 and 27 is $\sqrt[3]{216} = 6$.

Exam­ple 1#

Find the G.M. of 3 and 27, and of $\displaystyle \frac{1}{2}$ and 32.

$\sqrt{3 \times 27} = \sqrt{81} = 9$, and $\displaystyle \sqrt{\frac{1}{2} \times 32} = \sqrt{16} = 4$.

Insert­ing geo­met­ric means#

To insert $n$ geo­met­ric means between $a$ and $b$ means to find $G_1, G_2, \ldots, G_n$ such that

$$a, \ G_1, \ G_2, \ \ldots, \ G_n, \ b$$

is a G.P. This G.P. has $n + 2$ terms, so $b = ar^{n+1}$, and the com­mon ratio is

$$\displaystyle r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$$

Then $G_1 = ar$, $G_2 = ar^2$, and in gen­eral $G_k = ar^k$.

Exam­ple 2#

Insert 3 geo­met­ric means between 2 and 162.

$n = 3$, so $\displaystyle r^4 = \frac{162}{2} = 81$ and $r = 3$. The means are $6, 18, 54$, and the G.P. is $2, 6, 18, 54, 162$.

Exam­ple 3#

Insert 4 geo­met­ric means between 3 and 96.

$\displaystyle r^5 = \frac{96}{3} = 32$, so $r = 2$. The means are $6, 12, 24, 48$.

The prod­uct of the inserted means#

In Exam­ple 2 the prod­uct of the means is $6 \times 18 \times 54 = 5832$. The sin­gle G.M. of 2 and 162 is $\sqrt{324} = 18$, and $18^3 = 5832$. This is no acci­dent: by Prop­erty 2 of a G.P., the means pair off from the two ends ($6 \times 54$, and 18 in the mid­dle), each pair hav­ing prod­uct $ab$. So:

The prod­uct of $n$ geo­met­ric means inserted between $a$ and $b$ is $\left(\sqrt{ab}\right)^n$.

Exam­ple 4#

Find the prod­uct of 5 geo­met­ric means inserted between 2 and 18.

The sin­gle G.M. is $\sqrt{36} = 6$, so the prod­uct is $6^5 = 7776$.

A.M. and G.M. com­pared#

For 4 and 25 the A.M. is $\displaystyle \frac{29}{2} = 14.5$ and the G.M. is 10. For 3 and 27 the A.M. is 15 and the G.M. is 9. In both cases the A.M. is larger. This is always so.

For any two pos­i­tive num­bers $a$ and $b$, $\text{A.M.} \ge \text{G.M.}$, that is $\displaystyle \frac{a + b}{2} \ge \sqrt{ab}$, with equal­ity only when $a = b$.

Proof.

$$\displaystyle \frac{a + b}{2} - \sqrt{ab} = \frac{a - 2\sqrt{ab} + b}{2} = \frac{\left(\sqrt{a} - \sqrt{b}\right)^2}{2}$$

A square is never neg­a­tive, so the dif­fer­ence is at least 0. It equals 0 only when $\sqrt{a} = \sqrt{b}$, that is when $a = b$.

Exam­ple 5#

Show that for every pos­i­tive num­ber $x$, $\displaystyle x + \frac{1}{x} \ge 2$.

Apply A.M. $\ge$ G.M. to the pos­i­tive num­bers $x$ and $\displaystyle \frac{1}{x}$:

$$\displaystyle \frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} = 1 \implies x + \frac{1}{x} \ge 2$$

Equal­ity holds only when $\displaystyle x = \frac{1}{x}$, that is $x = 1$.

Find­ing two num­bers from their means#

If the A.M. of two num­bers is $A$ and the G.M. is $G$, then their sum is $2A$ and their prod­uct is $G^2$. So the num­bers are the roots of the qua­dratic equa­tion

$$x^2 - 2Ax + G^2 = 0$$

Exam­ple 6#

The A.M. of two num­bers is 10 and their G.M. is 8. Find the num­bers.

Their sum is 20 and their prod­uct is 64. So they are the roots of $x^2 - 20x + 64 = 0$, that is $(x - 4)(x - 16) = 0$. The num­bers are 4 and 16. Check: $\displaystyle \frac{4 + 16}{2} = 10$ and $\sqrt{64} = 8$.

Exam­ple 7#

If the G.M. of two num­bers is 12 and one of them is 9, find the other and the A.M. of the two.

$\sqrt{9b} = 12$ gives $9b = 144$, so $b = 16$. The A.M. is $\displaystyle \frac{9 + 16}{2} = 12.5$, which is indeed big­ger than 12.

Prac­tice#

1. Find the G.M. of 5 and 45.
2. Find the G.M. of $\displaystyle \frac{4}{9}$ and $\displaystyle \frac{9}{4}$.
3. Find the G.M. of 1, 9 and 81.
4. Insert 2 geo­met­ric means between 4 and 108.
5. Insert 3 geo­met­ric means between 1 and 256.
6. Insert 4 geo­met­ric means between 160 and 5.
7. Find the prod­uct of 4 geo­met­ric means inserted between 3 and 12.
8. The G.M. of two num­bers is 6 and one of them is 4. Find the other.
9. The A.M. of two num­bers is 25 and their G.M. is 15. Find the num­bers.
10. Show that for pos­i­tive $a$ and $b$, $\displaystyle \frac{a}{b} + \frac{b}{a} \ge 2$.

Answers#