Problem - 3319
Let real numbers $a$, $b$, and $c$ satisfy
$$
\left\{
\begin{array}{rcl}
a^2 - bc-8a +7&=&0\\
b^2 + c^2 +bc-6a+6&=&0
\end{array}
\right.
$$
Show that $1 \le a \le 9$.
The $1^{st}$ equation implies $bc=a^2 -8a+7=0$.
Subtracting the $2^{nd}$ equation from the $1^{st}$ one leads to
$$a^2 -2a +1 -(b^2+c^2-2bc)=0\implies b-c=\pm(a-1)$$
Therefore $b$ and $c$ are the two real roots of equation
$$x^2 \pm (a-1)x +(a^2-8a+7)=0$$
This means its determinant is non-negative, i.e. $$(a-1)^2 - 4\times (a^2-8a+7)\ge 0 \implies (a-1)(a-9)\le 0 \implies 1\le a\le 9$$