2. Determine the maximum possible value of the expression

$$
\frac{a_{1} a_{2} \ldots a_{n}}{\left(1+a_{1}\right)\left(a_{1}+a_{2}\right) \ldots\left(a_{n-1}+a_{n}\right)\left(a_{n}+2^{n+1}\right)}
$$

where $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are positive real numbers.
Solution. We seek the minimum possible value of the expression

$$
\frac{\left(1+a_{1}\right)\left(a_{1}+a_{2}\right) \ldots\left(a_{n-1}+a_{n}\right)\left(a_{n}+2^{n+1}\right)}{a_{1} a_{2} \ldots a_{n}}
$$

We have

$$
\frac{\left(1+a_{1}\right)\left(a_{1}+a_{2}\right) \ldots\left(a_{n-1}+a_{n}\right)\left(a_{n}+2^{n+1}\right)}{a_{1} a_{2} \ldots a_{n}}=\left(1+a_{1}\right)\left(1+\frac{a_{2}}{a_{1}}\right) \ldots\left(1+\frac{a_{n}}{a_{n-1}}\right)\left(1+\frac{2^{n+1}}{a_{n}}\right)
$$

We introduce the substitutions $x_{1}=a_{1}, x_{2}=\frac{a_{2}}{a_{1}}, \ldots, x_{n}=\frac{a_{n}}{a_{n-1}}, x_{n+1}=\frac{2^{n+1}}{a_{n}}$. Thus, we seek the minimum value of the expression

$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{n}\right)\left(1+x_{n+1}\right)
$$

where $x_{1} x_{2} \ldots x_{n} x_{n+1}=2^{n+1}$. Therefore,

$$
\begin{aligned}
&\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{n+1}\right)=1+x_{1}+x_{2}+\ldots+x_{n+1}+\sum_{\substack{k=1 \\
i<k}}^{n+1} x_{i} x_{k}+\ldots+x_{1} x_{2} \ldots x_{n+1} \\
& \geq 1+\binom{n+1}{1}^{n+1} \sqrt{x_{1} x_{2} \ldots x_{n+1}}+\binom{n+1}{2} \sqrt[\binom{n+1}{2}]{x_{1} x_{2} \ldots x_{n+1}}+x_{1} x_{2} \ldots x_{n+1} \\
&=\left(1+\sqrt[n+1]{x_{1} x_{2} \ldots x_{n+1}}\right)^{n+1}=\left(1+\sqrt[n+1]{2^{n+1}}\right)^{n+1}=3^{n+1}
\end{aligned}
$$

Accordingly, the minimum possible value of the expression (**) is $3^{n+1}$, and it is achieved when $x_{1}=x_{2}=\ldots=x_{n}=x_{n+1}=2$. This means that the maximum possible value of the expression (*) is $\frac{1}{3^{n+1}}$, and it is achieved when $a_{1}=2, a_{2}=2^{2}, a_{3}=2^{3}, \ldots, a_{n}=2^{n}$.