设函数f(x)=x(x-1)(x-a)(a>1),若x1,x2为方程f'(x)=0的两个实数根,且不等式f(x1)+f(

f'(x)=3x^2-2(a+1)x+a
若f'(x)=0则x^2=2(a+1)/3 x+a/3
f(x)=x(x^2-(a+1)x+a)=x(-(a+1)x+4a)/3=-2(a^2-4a+1)x/9-a(a+1)/9
f(x1)+f(x2)=-2(a^2-4a+1)(x1+x2)/9-2a(a+1)/9=-4(a+1)(a^2-4a+1)/27-2a(a+1)/9=0
由于a>1,于是a>2