7. (2004 National College Entrance Examination Chongqing Paper) Let the function $f(x)=x(x-1)(x-a)(a>1)$.
( I ) Find the derivative $f^{\prime}(x)$, and prove that $f(x)$ has two distinct extreme points $x_{1} 、 x_{2}$;
(II) If the inequality $f\left(x_{1}\right)+f\left(x_{2}\right) \leqslant 0$ holds, find the range of values for $a$.
7. ( I ) $f^{\prime}(x)=3 x^{2}-2(1+a) x+a$.

Let $f^{\prime}(x)=0$, we get $3 x^{2}-2(1+a) x+a=0$.
Since $\Delta=4\left(a^{2}-a+1\right)>0$, the equation (1) has two distinct real roots $x_{1} 、 x_{2}$.
Assume $x_{1}<x_{2}$, when $x<x_{1}$ or $x>x_{2}$, $f^{\prime}(x)>0$; when $x_{1}<x<x_{2}$, $f^{\prime}(x)<0$.
Therefore, $x_{1}$ is a point of local maximum, and $x_{2}$ is a point of local minimum.
( II ) From $f\left(x_{1}\right)+f\left(x_{2}\right) \leqslant 0$,
we have $x_{1}^{3}+x_{2}^{3}-(1+a)\left(x_{1}^{2}+x_{2}^{2}\right)+a\left(x_{1}+x_{2}\right) \leqslant 0$,
which is $\left(x_{1}+x_{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right]-(1+a)\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]+a\left(x_{1}+x_{2}\right) \leqslant 0$.
Also, $x_{1}+x_{2}=\frac{2}{3}(1+a), x_{1} x_{2}=\frac{a}{3}$,
substituting and dividing both sides by $1+a$, we get $2 a^{2}-5 a+2 \geqslant 0$.
Solving, we get $a \leqslant \frac{1}{2}$ (discard) or $a \geqslant 2$.
Therefore, when $a \geqslant 2$, the inequality $f\left(x_{1}\right)+f\left(x_{2}\right) \leqslant 0$ holds.