Question 69, The quadratic function $f(x)=x^{2}+m x+n$ has real roots, the inequality $s \leq(m-1)^{2}+(n-1)^{2}+$ $(m-n)^{2}$ holds for any quadratic function satisfying the above condition, then the maximum value of $s$ is $\qquad$ _.
Question 69, Solution Method One: In fact, the maximum value of the desired $s$ is $\min _{m^{2} \geq 4 n}\left\{(m-1)^{2}+(n-1)^{2}+\right.$ $\left.(m-n)^{2}\right\}$.

Notice that $(m-1)^{2}+(n-1)^{2}+(m-n)^{2}=\frac{9}{8}+\frac{(2 m-4 n-1)^{2}+6(m-1)^{2}+6\left(m^{2}-4 n\right)}{8} \geq \frac{9}{8}$, with equality when $m=$ $1, n=\frac{1}{4}$. Therefore, the maximum value of the desired $s$ is $\frac{9}{8}$.
Solution Method Two: Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{mx}+\mathrm{n}$ have two real roots $\mathrm{p}$ and $\mathrm{q}$, then $\left\{\begin{array}{c}p+q=-m \\ p q=n\end{array}\right.$, hence
$$
\begin{array}{l}
(m-1)^{2}+(n-1)^{2}+(m-n)^{2} \\
=(1+p+q)^{2}+(p q-1)^{2}+(p q+p+q)^{2} \\
=2\left(p^{2}+p+1\right)\left(q^{2}+q+1\right) \geq \frac{9}{8}
\end{array}
$$

Equality holds if and only if $p=q=-\frac{1}{2}$. Therefore, the maximum value of the desired $s$ is $\frac{9}{8}$.