# For what x an y is y / ( x^3 + 3 )^2 > 2/(x/y-y-3)?

Jul 11, 2016

See below

#### Explanation:

Compacting the inequality we have

-(y (18 - x + 12 x^3 + 2 x^6 + 3 y + y^2))/((3 + x^3)^2 (x - 3 y - y^2) )>0 or
$- \frac{y \left(18 - x + 12 {x}^{3} + 2 {x}^{6} + 3 y + {y}^{2}\right)}{x - 3 y - {y}^{2}} > 0$
because

${\left(3 + {x}^{3}\right)}^{2} \ge 0 \forall x \in \mathbb{R}$

The feasible set frontier is composed of

${f}_{1} \left(x , y\right) = - y = 0$
${f}_{2} \left(x , y\right) = 18 - x + 12 {x}^{3} + 2 {x}^{6} + 3 y + {y}^{2} = 0$

and

${f}_{3} \left(x , y\right) = x - 3 y - {y}^{2} = 0$

The feasible region interior is the set of points $\left\{x , y\right\}$ obeying

${f}_{1} \left(x , y\right) {f}_{2} \left(x , y\right) > 0 \mathmr{and} {f}_{3} \left(x , y\right) > 0$

or

${f}_{1} \left(x , y\right) {f}_{2} \left(x , y\right) < 0 \mathmr{and} {f}_{3} \left(x , y\right) < 0$

This set is shown in the attached figure in light blue