No Equality

We have $(xy)^2+2y+1=0$ $\Leftrightarrow[\dfrac{xy}{3y+1}\times(3y+1)]^2+2y+1=0$ We called $A=\dfrac{xy}{3y+1}$ So we get $A^2(3y+1)^2+2y+1=0$ $\Leftrightarrow9A^2y^2+(6A^2+2)y+A^2+1=0$ Set $A^2=a(a\geq0)$ $=>9ay^2+(6a+2)y+a+1=0$ $\Delta=(6a+2)^2-4\times9a(a+1)$ We get a real value of y when $\Delta\geq0$ $\Leftrightarrow(6a+2)^2-4\times9a(a+1)\geq0$ $\Leftrightarrow36a^2+12a+4-36a^2-36a\geq0$ $\Leftrightarrow4-12a\geq0$ $\Leftrightarrow a\leq\dfrac{1}{3}$ $\Rightarrow A^2\leq\dfrac{1}{3}$ $\Leftrightarrow -\sqrt{\dfrac{1}{3}}\leq A\leq\sqrt{\dfrac{1}{3}}$ $\Rightarrow A_{min}+A_{max} =0$