It is proved that the curve y = (x + 1) / (x ^ 2 + 1) has three inflection points on the same line After getting the second derivative of the function, I don't know how to get all three points

It is proved that the curve y = (x + 1) / (x ^ 2 + 1) has three inflection points on the same line After getting the second derivative of the function, I don't know how to get all three points

The first and second derivatives are obtained as follows:
y'=[(x²+1)-(x+1)(2x)]/[(x²+1)²]
=(-x²-2x+1)/[(x²+1)²]
y"=[(-2x-2)(x²+1)²-(-x²-2x+1)(4x)(x²+1)]/[(x²+1)^4]
=2(x³+3x²-3x-1)/[(x²+1)³]
=2(x-1)(x²+4x+1)/[(x²+1)³]
Let Y "= 0
x(1)=1,
x(2)=-2+√3,
x(3)=-2-√3
Consider the sign change law of Y "in the four intervals divided by the above solutions,
The above solutions are abscissa of inflection point,
Substituting into the analytic expression of Y, the ordinate of inflection point can be obtained as
y(1)=1,
y(2)=(1+√3)/4,
y(3)=(1-√3)/4
That is, the three inflection points of the curve y are
A(1,1),
B(-2+√3,(1+√3)/4),
C(-2-√3,(1-√3)/4)
Consider vectors,
AB=(-3+√3,(-3+√3)/4)=√3·(-√3+1,(-√3+1)/4)
BC=(-3-√3,(-3-√3)/4)=√3·(-√3-1,(-√3-1)/4)
Because the abscissa and ordinate of vectors AB and BC are proportional,
So ab ‖ BC, and B is the common point
So a, B and C are collinear