Example 7 Let real numbers $a$, $b$, $c$ satisfy
$$\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0, \\
b^{2}+c^{2}+b c-6 a+6=0
\end{array}\right.$$

Find the range of real number $a$.
$$\begin{array}{l}
\left\{\begin{array}{l}
b^{2}+c^{2}=-a^{2}+14 a-13, \\
2 b c=2 a^{2}-16 a+14
\end{array}\right. \\
\because \quad b^{2}+c^{2} \geqslant 2 b c, \\
\therefore \quad-a^{2}+14 a-13 \geqslant 2 a^{2}-16 a+14,
\end{array}$$

that is, $a^{2}-10 a+9 \leqslant 0$,
thus $1 \leqslant a \leqslant 9$.
To find the range of values for $a$, it is necessary to establish an inequality relationship for $a$. The arithmetic mean-geometric mean inequality, discriminant, and properties of non-negative numbers are commonly used tools.