Problem 1088

Proof (1) According to the problem, let $f(x)-x=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. When $x \in\left(0, x_{1}\right)$, since $00$, thus $a\left(x-x_{1}\right)\left(x-x_{2}\right)>0$, which means $f(x)>x$ holds.
Also, $x_{1}-f(x)=x_{1}-x-a\left(x-x_{1}\right)\left(x-x_{2}\right)=\left(x_{1}-x\right)\left[1+a\left(x-x_{2}\right)\right]$
Since $00,1+a\left(x-x_{2}\right)=1+a x-a x_{2}>1-a x_{2}>0$
we get $x_{1}>f(x)$
(2) Clearly, $x_{0}=-\frac{b}{2 a}$. By Vieta's formulas, $x_{1}$ and $x_{2}$ are the roots of the equation $f(x)-x=0$, i.e., $a x^{2}+(b-1) x+c=0$, so $x_{1}+x_{2}=-\frac{b-1}{a}, x_{0}=-\frac{b}{2 a}=\frac{a x_{1}+a x_{2}-1}{2 a}$. Since $a x_{2}<1$, it follows that $x_{0}<\frac{a x_{1}}{2 a}=\frac{x_{1}}{2}$