## Problem 6.

Find all real solutions of the equation

$$
3^{x^{2}-x-y}+3^{y^{2}-y-z}+3^{z^{2}-z-x}=1
$$
Solution. Applying the arithmetic mean-geometric mean inequality results in

$$
\begin{aligned}
& 3^{x^{2}-x-y}+3^{y^{2}-y-z}+3^{z^{2}-x-x} \geq 3 \sqrt[3]{3^{x^{2}+y^{2}+z^{2}-2 x-2 y-2 z}}=3^{\frac{1}{3}\left(x^{2}+y^{2}+z^{2}-2 x-2 y-2 z\right)+1} \\
& =3^{\frac{1}{3}\left[(x-1)^{2}+(y-1)^{2}+(z-1)^{2}\right]} \geq 3^{0}=1
\end{aligned}
$$

Equality holds when $x=y=z=1$. Therefore, the only solution is $(x, y, z)=(1,1,1)$ and we are done.