How to show that $\left| \frac{-x^2y-y^3+y}{x^2+y^2} \right| < \pi$?

Question

How to show that

$$\left| \frac{-x^2y-y^3+y}{x^2+y^2} \right| < \pi$$

if $1<\frac{1}{x^2+y^2}<4$?

I think of multiplying $-x^2y-y^3+y$ to the given condition, but it could further complicate things. Is there another way to have a more simplified inequality?

Answer 1

First look at the numerator-$$-x^2y-y^3+y=-(x^2+y^2)y+y=[-(x^2+y^2)+1]y.$$ Now, bring the condition to that form:

$1<\frac{1}{x^2+y^2}<4\quad\quad\quad\dots(0)$
$\Rightarrow1>x^2+y^2>\frac{1}{4}\quad\quad\quad\dots(1)$
$\Rightarrow-1<-(x^2+y^2)<-\frac{1}{4}$
$\Rightarrow0<-(x^2+y^2)+1<\frac{3}{4}$
$\Rightarrow\left|-(x^2+y^2)+1\right|<\frac{3}{4}\quad\quad\quad\dots(2)$

Also, since $x^2\ge0$ (since square of a real number) and $x^2+y^2<1$ from (1), we get $$0<y^2<1\Rightarrow0<|y|<1\quad\quad\quad\dots(3)$$ Multiplying (2) by (3), $$\left|-(x^2+y^2)+1\right|\cdot|y|<\frac{3}{4}\cdot1\Rightarrow\left|\left[-(x^2+y^2)+1\right]y\right|=\left|-x^2y-y^3+y\right|<\frac{3}{4}~\dots(4)$$ Multiplying $(0)$ with (4) (since, $1/(x^2+y^2)$ is a positive quantity, it is equal to $|1/(x^2+y^2)|$), we get $$\left|-x^2y-y^3+y\right|\cdot\left|\frac{1}{x^2+y^2}\right|<\frac{3}{4}\cdot4\Rightarrow\left|\frac{-x^2y-y^3+y}{x^2+y^2}\right|<3<\pi$$

Answer 2

$\dfrac{-x^2y-y^3+y}{x^2+y^2} = \dfrac{y-y(x^2+y^2)}{x^2+y^2}= y\left(\dfrac{1}{x^2+y^2} - 1\right) $

$0 \lt \left(\dfrac{1}{x^2+y^2} - 1\right) \lt 3 $

$1/4 \lt x^2+y^2 \lt 1 \implies 0 \leq |y| \lt 1$

From here its easy to get the desired inequality since $3 \lt \pi$

Answer 3

Another way, by polar coordinates with $\frac12<r<1$

$$\left| \frac{-x^2y-y^3+y}{x^2+y^2} \right|=\left|\frac{-r^3\sin \theta+r\sin \theta}{r^2}\right| =|\sin \theta|\cdot \frac{1-r^2}{r}<\frac1r-r<2-\frac12=\frac{3}2<\pi$$

therefore the stronger inequality holds

$$\left| \frac{-x^2y-y^3+y}{x^2+y^2} \right|<\frac 3 2 <\frac \pi 2<\pi$$