So 0

y=(a^2+3)/a+b^2/(b+ 1)

=(a^2+3)/a+( 1-a)^2/(2-a)，

Multiply both sides by a(2-a) and you get

(2a-a^2)y=(a^2+3)(2-a)+a( 1-a)^2

=6-3a+2a^2-a^3

.....+a-2a^2+a^3

=6-2a，

Ya 2-a (2+2y)+6 = 0，

a，y∈R，

So △/4 = (1+y) 2-6y = y 2-4y+1≥ 0,

So y≤2-√3 or y≥2+√3,

When a=( 1+y)/y=3+√3 (house) or 3-√3, take the equal sign.

So the minimum value of y is 2+√3, which is what you want.

Solution 2 y=(a? +3)/a+b？ /(b+ 1)

= a+3/a+(b- 1)+ 1/(b+ 1)

=3/a+ 1/(b+ 1)

=3/a+ 1/(2-a)

=(6-2a)/[2a-a^2)，

Let u=3-a∈( 1, 3), then a=3-u,

y=2u/[2(3-u)-(3-u)^2]

=2u/(-3+4u-u^2)

=2/[4-(3/u+u)]

3/u+u≥2√3, when u=√3, take the equal sign,

So 4-(3/u+u) ∈ (0,4-2 √ 3),

So y≥2/(4-2√3)=2+√3,

So the minimum value of y is 2+√3.