Proof of quadratic inequality using AM-GM

Question

Proof of quadratic inequality using AM-GM

Answer 1

Hint: Expand $(xy-2)^2\ge0$ and $(x-y)^2\ge0$.

Answer 2

$x^2y^2+x^2+y^2+4-6xy=x^2y^2+(x-y)^2+4-4xy=(xy-2)^2+(x-y)^2\geq 0.$

Answer 3

we have to show that $$x^2y^2+x^2+y^2+4-6xy\geq 0$$ dividing by $x^2+1$ gives $$y^2-\frac{6xy}{x^2+1}+\frac{4+x^2}{x^2+1}\geq 0$$ this is equivalent to $$\left(y-\frac{3x}{x^2+1}\right)^2+\frac{(4+x^2)(x^2+1)-9x^2}{(x^2+1)^2}\geq 0$$ and this is equivalent to $$\left(y-\frac{3x}{x^2+1}\right)^{ 2 }+\frac{(x^2-2)^2}{(x^2+1)^2}\geq 0$$ which is true.