Solve the following system of equations on the set of real numbers:

$$
\begin{gathered}
\log _{3} a+\log _{3} b+\log _{3} c=0 \\
3^{3^{a}}+3^{3^{b}}+3^{3^{c}}=81
\end{gathered}
$$
From the first equation, it follows that $\log _{3} a b c=0$, which means $a b c=1$, and also that $a, b, c$ are positive. We will show that for every such $a, b, c$,

$$
3^{3^{a}}+3^{3^{b}}+3^{3^{c}} \geq 81
$$

and equality holds only when $a=b=c$. By the arithmetic and geometric mean inequality, with the notation $a+b+c=d$,

$$
\begin{gathered}
\frac{3^{3^{a}}+3^{3^{b}}+3^{3^{c}}}{3} \geq \sqrt[3]{3^{3^{a}} \cdot 3^{3^{b}} \cdot 3^{3^{c}}}=3^{\left(3^{a}+3^{b}+3^{c}\right) / 3} \geq \\
\geq 3^{\sqrt[3]{3^{a} \cdot 3^{b} \cdot 3^{c}}}=3^{\sqrt[3]{3^{a+b+c}}}=3^{3^{\frac{d}{3}}}
\end{gathered}
$$

(In the second inequality, we used the fact that if the exponent $x$ decreases, then $3^{x}$ also decreases.) Finally,

$$
\frac{d}{3}=\frac{a+b+c}{3} \geq \sqrt[3]{a b c}=1, \text { so } 3^{3^{d / 3}} \geq 3^{3}=27
$$

This gives us that $\left(3^{3^{a}}+3^{3^{b}}+3^{3^{c}}\right) / 3 \geq 27$, from which (1) follows. If we review the inequalities, we always used the arithmetic and geometric mean inequality, as well as the (strict) monotonicity of the function $x \mapsto 3^{x}$, so equality holds everywhere if and only if $a=b=c$, and by the assumption $a b c=1$, this happens when $a=b=c=1$. This is the only and indeed correct solution to the equation.

Gergely Harcos (Bp., ELTE Apáczai Csere J. Gymn., III. o. t.)