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

Jul 3, 2016

There is no pair $\left\{x , y\right\}$ satisfying the condition.

#### Explanation:

The pairs $\left\{x , y\right\}$ satisfying

$\frac{y}{x + 3} ^ 2 > \frac{2}{x - y - 3}$

also satisfy

$y \left(x - y - 3\right) > 2 {\left(x + 3\right)}^{2}$

and also satisfy

$y \left(x - y - 3\right) - 2 {\left(x + 3\right)}^{2} > 0$

Calling now

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

let us find the local minima/maxima: Solving the conditions

$\nabla f \left(x , y\right) = \vec{0}$

or

{ (12 - 4 x + y = 0), (-3 + x - 2 y = 0) :}

with solution

$\left\{x = 3 , y = 0\right\}$

for $f \left(x , y\right)$ we have

${\nabla}^{2} f \left(x , y\right) = \left(\begin{matrix}- 4 & 1 \\ 1 & - 2\end{matrix}\right)$ with eigenvalues

$\left\{- 3 - \sqrt{2} , - 3 + \sqrt{2}\right\}$ so $f \left(x , y\right)$ has a maximum which is a global maximum, at $\left\{x = 3 , y = 0\right\}$.

Evaluating $f \left(3 , 0\right) = 0$ we see that there is no pair $\left\{x , y\right\}$ which obeys strictly the condition.