If three positive numbers a,b,c are in A.P.

Question

If three positive numbers $a,b,c$ are in A.P. such that $abc=8$, then the minimum possible value of $b$ is:

Accepted Answer

Since $a,b,c$ are in A.P., $a+c=2b$. By AM--GM on $a,c$: $a+c\ge 2\sqrt{ac}$, so $2b\ge2\sqrt{ac}=2\sqrt{8/b}$ (using $ac=8/b$).

$$b\ge\sqrt{8/b}\ \Rightarrow\ b^2\ge\frac8b\ \Rightarrow\ b^3\ge8\ \Rightarrow\ b\ge2.$$

Equality holds at $a=b=c=2$ (indeed $abc=8$). So the minimum value of $b$ is $2$.

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