Geometric Mean and the A.M.-G.M. Inequality

The Geometric Mean

Key Point (Definition): The geometric mean of two positive numbers $a$ and $b$ is $G = \sqrt{ab}$ — the number that makes $a, G, b$ a G.P. The G.M. of 2 and 8 is 4, and indeed $2, 4, 8$ has constant ratio 2.

Inserting $n$ geometric means: to place $G_1, G_2, \ldots, G_n$ between positive $a$ and $b$ so that $a, G_1, \ldots, G_n, b$ is a G.P., note $b$ is the $(n+2)$-th term, so $b = ar^{n+1}$:

$$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}, \qquad G_k = ar^k = a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}$$

For example: three means between 1 and 256 need $256 = r^4$, so $r = \pm 4$ — insert $4, 16, 64$ (or $-4, 16, -64$). Likewise, two means between 3 and 81 need $r^3 = 27$: insert 9 and 27.

[Board Tip] Count carefully: inserting $n$ means makes $n + 2$ terms, so the exponent on $r$ is $n + 1$ — the most common slip in these problems.

A.M. ≥ G.M.

Let $A = \frac{a+b}{2}$ and $G = \sqrt{ab}$ for positive reals $a, b$. Then

$$A - G = \frac{a + b - 2\sqrt{ab}}{2} = \frac{\left(\sqrt{a} - \sqrt{b}\right)^2}{2} \ge 0$$

Key Point: $A \ge G$ always, with equality exactly when $a = b$ — the square vanishes only then.

Recovering the numbers from $A$ and $G$: $A = 10, G = 8$ give $a + b = 20$ and $ab = 64$. Then $(a-b)^2 = (a+b)^2 - 4ab = 400 - 256 = 144$, so $a - b = \pm 12$: the numbers are 4 and 16.

The general recovery — the numbers are $A \pm \sqrt{(A+G)(A-G)}$ — is real precisely because $A \ge G$.

[JEE Tip] The identity $(a-b)^2 = (a+b)^2 - 4ab$ is the bridge from (sum, product) to the pair itself — the same bridge used for quadratic roots. A.M./G.M. of the roots of $x^2 - px + q = 0$ are $\frac{p}{2}$ and $\sqrt{q}$; given A.M. 8 and G.M. 5, the equation is $x^2 - 16x + 25 = 0$.

Solved Examples

Example 1: Insert three means

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.

Solution:

Step 1 — Count the terms. Three insertions make 5 terms, so 256 is the 5th: $256 = 1 \cdot r^4$.

Step 2 — Solve for $r$. $r^4 = 256$ gives $r = \pm 4$.

Step 3 — Write the means. For $r = 4$: $G_1 = 4$, $G_2 = 16$, $G_3 = 64$. For $r = -4$: $-4, 16, -64$.

Step 4 — Check. $1, 4, 16, 64, 256$ has constant ratio 4 ✓.

Takeaway: An even power of $r$ always admits both signs — report both G.P.s unless positivity is demanded.

Example 2: Insert two means

Insert two numbers between 3 and 81 so that the resulting sequence is a G.P.

Solution:

Step 1 — Count. Two insertions make 4 terms: $81 = 3r^3$.

Step 2 — Solve. $r^3 = 27$ gives $r = 3$ (odd power — one real root).

Step 3 — Write the means. $9$ and $27$: the G.P. is $3, 9, 27, 81$ ✓.

Takeaway: Odd exponent → a single ratio; the sign ambiguity of Example 1 disappears.

Example 3: Numbers from A.M. and G.M.

If the A.M. and G.M. of two positive numbers are 10 and 8, find the numbers.

Solution:

Step 1 — Convert to sum and product. $a + b = 2A = 20$; $ab = G^2 = 64$.

Step 2 — Bridge to the difference. $(a-b)^2 = (a+b)^2 - 4ab = 400 - 256 = 144$, so $a - b = \pm 12$.

Step 3 — Solve the pair. $a + b = 20$, $a - b = 12$: $a = 16$, $b = 4$.

Step 4 — Check. A.M. $= \frac{20}{2} = 10$ ✓; G.M. $= \sqrt{64} = 8$ ✓.

Takeaway: $(a-b)^2 = (a+b)^2 - 4ab$ is the bridge from (sum, product) to the numbers themselves.

Example 4: The G.M. of two specific numbers

Find the geometric mean of (i) 4 and 9 (ii) $\frac{1}{2}$ and 8.

Solution:

Step 1 — (i). $G = \sqrt{4 \times 9} = \sqrt{36} = 6$.

Step 2 — (ii). $G = \sqrt{\frac{1}{2} \times 8} = \sqrt{4} = 2$.

Takeaway: The G.M. tames mismatched sizes — a half and an 8 average (geometrically) to 2.

Example 5: Quadratic from A.M. and G.M.

The A.M. and G.M. of the roots of a quadratic are 8 and 5. Find the equation.

Solution:

Step 1 — Convert. Sum of roots $= 2 \times 8 = 16$; product $= 5^2 = 25$.

Step 2 — Assemble. $x^2 - (\text{sum})x + (\text{product}) = 0$: $x^2 - 16x + 25 = 0$.

Takeaway: A.M. and G.M. are just disguised sum and product — quadratics follow immediately.

Example 6: An exponent from the G.M.

Find $n$ so that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n}$ is the geometric mean of $a$ and $b$.

Solution:

Step 1 — Set the equation. $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = a^{1/2}b^{1/2}$; cross-multiply:

$$a^{n+1} + b^{n+1} = a^{n+\frac{1}{2}}b^{\frac{1}{2}} + a^{\frac{1}{2}}b^{n+\frac{1}{2}}$$

Step 2 — Group and factor. $a^{n+\frac{1}{2}}\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right) = b^{n+\frac{1}{2}}\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right)$.

Step 3 — Cancel and conclude. For $a \ne b$: $\left(\frac{a}{b}\right)^{n+\frac{1}{2}} = 1$, forcing $n + \frac{1}{2} = 0$: $n = -\frac{1}{2}$.

Takeaway: Collect the powers of $a$ on one side and $b$ on the other — a ratio-equals-1 equation kills the exponent.

Example 7: Ratio from a sum condition

The sum of two numbers is 6 times their geometric mean. Show the numbers are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$.

Solution:

Step 1 — Normalise by the G.M. $a + b = 6\sqrt{ab}$; divide by $\sqrt{ab}$ and set $t = \sqrt{\frac{a}{b}}$: $t + \frac{1}{t} = 6$.

Step 2 — Solve the quadratic in $t$. $t^2 - 6t + 1 = 0$ gives $t = 3 \pm 2\sqrt{2}$.

Step 3 — Convert back to the ratio. $\frac{a}{b} = t^2$; with $t = 3 + 2\sqrt{2}$, note $\frac{1}{t} = 3 - 2\sqrt{2}$, so

$$\frac{a}{b} = \frac{t}{1/t} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$$

— exactly the stated ratio. ∎

Takeaway: The substitution $t = \sqrt{a/b}$ converts symmetric sum-G.M. conditions into one quadratic.

Example 8: Ratio of A.M. to G.M.

If the ratio of the A.M. and G.M. of two positive numbers is $m : n$, show $a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)$.

Solution:

Step 1 — Set up. $\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}$. With $t = \sqrt{\frac{a}{b}}$: $t + \frac{1}{t} = \frac{2m}{n}$.

Step 2 — Solve. $nt^2 - 2mt + n = 0$ gives $t = \frac{m + \sqrt{m^2 - n^2}}{n}$ (taking $a \ge b$).

Step 3 — Form the ratio. The other root is $\frac{m - \sqrt{m^2 - n^2}}{n}$ and the two roots multiply to 1, so $\frac{a}{b} = t^2 = \frac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}}$. ∎

Takeaway: The same $t$-substitution generalises Example 7 — with $m : n = 3 : 1$ it reproduces it exactly.

Example 9: Equality case of A.M.-G.M.

For which positive pairs $(a, b)$ does $\frac{a+b}{2} = \sqrt{ab}$ hold?

Solution:

Step 1 — Use the identity. $A - G = \frac{(\sqrt{a} - \sqrt{b})^2}{2}$.

Step 2 — Set it to zero. A square vanishes only when its base does: $\sqrt{a} = \sqrt{b}$.

Step 3 — Conclude. Exactly the pairs with $a = b$.

Takeaway: The equality case is as important as the inequality — optimisation arguments hinge on it.

Example 10: A quick A.M.-G.M. application

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

Solution:

Step 1 — Choose the pair. Apply $A \ge G$ to $a = x$, $b = \frac{1}{x}$.

Step 2 — Compute the G.M. $\sqrt{x \cdot \frac{1}{x}} = 1$ — the product is engineered to be constant.

Step 3 — Conclude. $\frac{x + \frac{1}{x}}{2} \ge 1$, so $x + \frac{1}{x} \ge 2$, with equality at $x = 1$.

Takeaway: A.M.-G.M. gives minimum-value results whenever a product stays constant — a JEE staple.