设f(x)=ax^2+bx+c当x=1+x^2/2即x=1时必有x=f(x)=1+x^2/2(不等式的性质)所以f(1)=1 f(-1)=0a+b+c=1a-b+c=0c=1/2-ab=1/2
则f(x)=ax^2+1/2x+1/2-a由x≤f(x)≤1+x^2/2 ax^2-1/2x+1/2-a>=0(a-1/2)x^2+1/2x-(a+1/2)<=0
f(x)=1/4x^2+1/2x+1/4
