5. Find all triples of real numbers $a, b, c$, for which

$$
27^{a^{2}+b+c+1}+27^{b^{2}+c+a+1}+27^{c^{2}+a+b+1}=3
$$
Solution. By the inequality of means

$$
\begin{gathered}
\frac{27^{a^{2}+b+c+1}+27^{b^{2}+c+a+1}+27^{c^{2}+a+b+1}}{3} \geqslant\left(27^{a^{2}+b+c+1} \cdot 27^{b^{2}+c+a+1} \cdot 27^{c^{2}+a+b+1}\right)^{1 / 3}= \\
=3^{a^{2}+b+c+1+b^{2}+a+c+1+c^{2}+a+b+1}=3^{(a+1)^{2}+(b+1)^{2}+(c+1)^{2}} \geqslant 1
\end{gathered}
$$

with equality holding only when $a=b=c=-1$.