请高手帮助解决一道高中数学题，这里先谢谢

y=ax∧2＋bx+c的图像经过点（-1，0）
0=a-b+c,b=a+c
y<=0.5(1+x^2)
ax^2+bx+c-1/2-1/2x^2<=0
(a-1/2)x^2+bx+c-1/2<=0 对任意x都成立
当a-1/2=0,a=1/2时，bx+c-1/2<=0,则b=0,c=-1/2

y=1/2x^2-1/2

当a-1/2≠0时.a<0,△=b^2-4*(a-1/2)(c-1/2)=(a-c)^2+2(a+c)-1<=0
x≤y对任意x都成立
ax^2+(b-1)x+c>=0 ,a>0,△=(b-1)^2-4*ac=(a-c)^2-2(a+c)+1<=0
两个不等式相加，2（a-c)^2<=0,a=c

ax^2+bx+c-1/2-1/2x^2<=0 代入x=1 ，a+b+c-1<=0
ax^2+(b-1)x+c>=0,代入x=1, a+b+c-1>=0
所以a+b+c-1=0,且b=a+c，a=c
得，a=c=1/4,b=1/2

y=1/4x^2+1/2x+1/4