Equation To Inequality

To solve this problem, you have to make $P$ appeared in the equation $(x^2-y^2+1)^2+4x^2y^2-x^2-y^2=0$ $\Leftrightarrow x^4-2x^2(y^2-1)+(y^2-1)^2+4x^2y^2-x^2-y^2=0$ $\Leftrightarrow x^4+y^4-2x^2y^2+4x^2y^2+2x^2-2y^2-x^2-y^2+1=0$ $\Leftrightarrow (x^2+y^2)^2-3(x^2+y^2)+1=-4x^2\leq 0;\forall x\in\mathbb{R}$ $\Rightarrow P^2-3P+1\leq 0$ $\Leftrightarrow\frac{3-\sqrt{5}}{2}\leq P\leq\frac{3+\sqrt{5}}{2}$ So the sum is $3.00$

If we expand everything in the given equation, we get:

$x^4+2x^2y^2+2x^2+y^4-2y^2+1-x^2-y^2=0$

$x^4+(2y^2+1)x^2+(y^4-3y^2+1)=0$ .

Using quadratic formula to solve for $x^2$ , as well as the fact that $x^2$ must be positive, we get:

$0\leq x^2=\frac{-2y^2-1\pm \sqrt{(2y^2+1)^2-4(y^4-3y^2+1)}}{2}=\pm \sqrt{4y^2-\frac{3}{4}}-y^2-\frac{1}{2}$ .

To this end, the right-hand side must be positive, so we only take the positive square root. Moving on, we have:

$0\leq \sqrt{4y^2-\frac{3}{4}}-y^2-\frac{1}{2}$

$y^2+\frac{1}{2}\leq \sqrt{4y^2-\frac{3}{4}}$

$y^4+y^2+\frac{1}{4}\leq 4y^2-\frac{3}{4}$

$y^4-3y^2+1\leq 0$ .

Once again using the quadratic formula for $y^2$ , and knowing that a quadratic in $y^2$ will be negative between the two roots, we get:

$\frac{3-\sqrt{5}}{2}\leq y^2\leq \frac{3+\sqrt{5}}{2}$ .

From here, by substituting for $x^2$ , note that $P=x^2+y^2=\Big( \sqrt{4y^2-\frac{3}{4}}-y^2-\frac{1}{2}\Big) +y^2=\sqrt{4y^2-\frac{3}{4}}-\frac{1}{2}$ .

Since this function $P$ is strictly increasing with regard to $y^2$ , we can say that the minimum and maximum of $P$ occur at the minimum and maximum values of $y^2$ . So:

$min(P)=\sqrt{6-2\sqrt{5}-\frac{3}{4}}-\frac{1}{2}=\frac{\sqrt{21-8\sqrt{5}}}{2}-\frac{1}{2}=\frac{4-\sqrt{5}}{2}-\frac{1}{2}$ $max(P)=\sqrt{6+2\sqrt{5}-\frac{3}{4}}-\frac{1}{2}=\frac{\sqrt{21+8\sqrt{5}}}{2}-\frac{1}{2}=\frac{4+\sqrt{5}}{2}-\frac{1}{2}$ .

And thus, $min(P)+max(P)=\frac{(4+\sqrt{5})+(4-\sqrt{5})}{2}-1=\boxed{3}$ .