If Three Positive Real Numbers a, b, c Are in AP Such That abc = 4, Then the Minimum Value of b Is

A standard AM-GM optimisation problem involving an arithmetic progression, not tied to a specific exam paper.

Problem

If three positive real numbers $a,b,c$ are in AP such that $abc=4$, then the minimum value of $b$ is?

Key insight. Because $a,b,c$ are in AP, their sum $a+b+c$ is exactly $3b$ — the middle term times three. That turns the AM-GM inequality on three numbers, which normally involves all of $a$, $b$, and $c$ separately, into an inequality purely about $b$ and the fixed product $abc$.

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Approach

Use the AP condition to write $a+c=2b$, so $a+b+c=3b$. Apply the AM-GM inequality to the three positive numbers $a,b,c$, substituting $a+b+c=3b$ and $abc=4$, to get a direct lower bound on $b$. Check that this bound is achieved when $a=b=c$.

Solution

Step 1 — Use the AP condition

Since $a,b,c$ are in AP, the middle term is the average of the outer two:

$$b = \frac{a+c}{2} \implies a+c=2b$$

Step 2 — Find a + b + c

$$a+b+c = (a+c)+b = 2b+b = 3b$$

Step 3 — Apply AM-GM to a, b, c

$$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$$

Step 4 — Substitute known values

$$\frac{3b}{3} \geq \sqrt[3]{4} \implies b\geq 4^{1/3} = 2^{2/3}$$

Step 5 — Confirm the bound is achievable

Equality in AM-GM holds when $a=b=c$. If $a=b=c$ and $abc=4$, then $a^3=4$, so $a=4^{1/3}=2^{2/3}$ — consistent with $b=2^{2/3}$, and $a=b=c$ trivially satisfies the AP condition. So the minimum is genuinely attained, not just a bound.

Answer

Common mistakes

• Forgetting to check that the AM-GM bound is actually achievable under the given constraints. AM-GM always gives a valid lower bound, but the minimum is only meaningful if some choice of $a,b,c$ satisfying all the given conditions actually attains it — here, $a=b=c=2^{2/3}$ does.
• Trying to minimise $b$ using calculus on two independent variables. Since $a,c$ are tied together by both the AP condition and the product condition, this is really a single-variable optimisation in disguise (via AM-GM) — setting up a two-variable Lagrange-style calculus approach is unnecessarily complicated here.

Practise next

• Three positive reals $p,q,r$ are in AP such that $pqr=27$. Find the minimum value of $q$, using the same AM-GM approach.