f(x)=x判别式1-4ab≥0
f[f(x)]=x中,令f(x)=y,则f(y)=x.
代入相减a(x-y)(x+y)=y-x,所以y=x或a(x+y)=-1
y=x即f(x)=x.
a(x+y)=-1及f(x)=y得ax^2+x+1/a+b=0,1-4a(1/a+b)≥0得ab≤-3/4
a^2+b^2≥2|ab|≥3/2.