Problem 3. Six numbers are written in a row. It is known that among them there is a one and any three adjacent numbers have the same arithmetic mean. Find the maximum value of the geometric mean of any three adjacent numbers in this row, if the arithmetic mean of all 6 numbers is A.
Solution. Let's write in a row $a_{1}, \ldots, a_{6}$. From the conditions

\[
\begin{aligned}
& (1 / 3)\left(a_{1}+a_{2}+a_{3}\right)=(1 / 3)\left(a_{2}+a_{3}+a_{4}\right) \\
& (1 / 3)\left(a_{2}+a_{3}+a_{4}\right)=(1 / 3)\left(a_{3}+a_{4}+a_{5}\right) \\
& (1 / 3)\left(a_{3}+a_{4}+a_{5}\right)=(1 / 3)\left(a_{4}+a_{5}+a_{6}\right)
\end{aligned}
\]

we get $a_{1}=a_{4}=x, a_{2}=a_{5}=y, a_{3}=a_{6}=z$. The row can be written as

\[
x, y, z, x, y, z
\]

Among the 6 numbers, there are at least three pairs of equal numbers. From (1) it follows that it does not matter which of the values is equal to 1, without loss of generality, let $z=1$ and rewrite (1) as $x, y, 1, x, y, 1$. From the condition on their arithmetic mean, we have $(1 / 6)(x+y+1+x+y+1)=(1 / 3)(x+y+1)=A$, from which

\[
y=3 A-1-x
\]

Let $p=3 A-1$, then $y=p-x$ and the geometric mean of three consecutive numbers is expressed as $g(x)=\sqrt[3]{x(p-x)}$. The function $g(x)$ has a maximum at the same point as the function

\[
f(x)=g^{3}(x)=x(p-x)
\]

This is a parabola, its maximum is achieved at the vertex with coordinates $\left(p / 2, p^{2} / 4\right)$. Thus, the maximum value of the function $g(x)=\sqrt[3]{f(x)}$ is

\[
\sqrt[3]{p^{2} / 4}=\sqrt[3]{(3 A-1)^{2} / 4}
\]

Answer: $\sqrt[3]{(3 A-1)^{2} / 4}$.