请教一个极值问题

已知函数f(x)=x^2+2x+1/(x^2)+a/x+b，当f(x)=0方程有解,求a^2+b^2的最小值
解：因为：f(x)=x^2+2x+1/(x^2)+a/x+b，f(x)=0方程有解，
所以：存在x，x不是0，x^2+2x+1/x^2+a/x+b=0
则：x^4+2x^3+1+ax+bx^2=0
x^4+2x^3+bx^2+ax+1=0
原方程有四个不为零的解可以写成
（x-x1)(x-x2)(x-x3)(x-x4)=0
即：[x^2-(x1+x2)x+x1x2][x^2-(x3+x4)x+x3x4]=0
x^4-(x1+x2)x^3+x1x2x^2-
(x3+x4)x^3+(x1+x2)(x3+x4)x^2-x1x2(x3+x4)x+
x3x4x^2-(x1+x2)x3x4x+x1x2x3x4=0
x^4-(x1+x2+x3+x4)x^3+[(x1+x2)(x3+x4)+x1x2+x3x4]x^2+[-x1x2(x3+x4)-(x1+x2)]x+x1x2x3x4=0
所以：a=-x1x2(x3+x4)-(x1+x2)x3x4
b=(x1+x2)(x3+x4)+x1x2+x3x4
x1x2x3x4=1
a^2+b^2=[x1x2(x3+x4)+(x1+x2)x3x4]^2+[(x1+x2)(x3+x4)+x1x2+x3x4]^2
=(x1x2x3+x2x3x4+x1x3x4+x1x2x4)^2+(x1x3+x1x4+x2x3+x2x4+x1x2+x3x4)^2
=(1/x4+1/x3+1/x2+1/x1)^2+(1/x2x4+1/x1x3+1/x1x4+1/x1x2+1/x2x3+1/x3x4)^2=[(x1+x2+x3+x4)/x1x2x3x4]^2+(x1x3+x1x4+x2x3+x2x4+x1x2+x3x4)^2=2^2+(x1x3+x1x4+x2x3+x2x4+x1x2+x3x4)^2
当后面一项为0时，a^2+b^2=4为最小值。