Given $x^2-y^2+2x+2y \geq 2xy+1$ and $y^2-x^2=\frac{1}{3}$ with the condition $x,y > 0$ find $\frac{x}{y}$ Had this question popup while studying for a test but unsure how to go about solving it:


Given $x^2-y^2+2x+2y \geq 2xy+1$ and $y^2-x^2=\frac{1}{3}$ with the condition $x,y > 0$ find $\frac{x}{y}$


Initial thoughts: It feels like $\frac{x}{y}$ doesn't have an actual value but a region but not completely sure.
This is my attempt at finding out $\frac{x}{y}$:
From the first condition
$$x^2-y^2+2x+2y \geq 2xy+1$$
$$ -(y^2-x^2)+2(x+y) \geq 2xy+1$$
$$ -\frac{1}{3}+2(x+y) \geq 2xy+1$$
$$ x+y-xy \geq \frac{2}{3} $$
$$ y(1-x) \geq \frac{2}{3}-x$$
$$ y(1-x) \geq \frac{2-3x}{3}$$
$$ y \geq \frac{2-3x}{3(1-x)} ~~~~\text{if x<1}$$
$$ y \leq \frac{2-3x}{3(1-x)} ~~~~\text{if x>1}$$
But stuck now :(

First, lets solve for $y$ in $y^2-x^2=\frac{1}{3}$. Since $y>0$, we find $$y=\sqrt{\frac{1+3x^2}{3}}.$$ Now define $g(x,y)=x^2-y^2+2x+2y-2xy-1$. As you noted, we can substitute to get $$g(x,y)=-\frac{1}{3}+2x+2y-2xy-1.$$ However, we can also get rid of $y$ with the previous condition: define $$f(x)=g(x,\sqrt{\frac{1+3x^2}{3}})=-\frac{1}{3}+2x+2\sqrt{\frac{1+3x^2}{3}}-2x\sqrt{\frac{1+3x^2}{3}}-1.$$ Now, the question is for what values of $x$ is $f(x)\geq 0$? Using mathematica and the fact that $x>0$, we find that the roots are $$x_1=\frac{1-\sqrt{\frac{7-4\sqrt{2}}{3}}}{2},\ x_2=\frac{1+\sqrt{\frac{7+4\sqrt{2}}{3}}}{2}.$$ Now, substituting into $$\frac{x}{y}=\frac{x}{\sqrt{\frac{1+3x^2}{3}}}$$ we get that $$\frac{x}{y}\in [\frac{3-\sqrt{21-12\sqrt{2}}}{\sqrt{6(7-2\sqrt{2}-\sqrt{21-12\sqrt{2}})}},\frac{3+\sqrt{21+12\sqrt{2}}}{\sqrt{6(7+2\sqrt{2}+\sqrt{21+12\sqrt{2}})}}]\approx[0.2754,0.9353]$$
