Inequality: $x-\sqrt{x^2+4y}<0 $

Question

I´m trying to find the points in the plane such that:

1) $x-\sqrt{x^2+4y}<0 $ and

2)$x+\sqrt{x^2+4y}>0$

I tried to divide it by cases: For the first inequality: if $x>0$ then $x<\sqrt{x^2+4y} \Rightarrow x^2 < x^2+4y \Rightarrow y>0$ The problem comes when I consider $x<0$ because I can´t guarantee that $x^2 <x^2+4y$.

I can plot the graph of the region but I´m looking for an analytic proof of the inequality.

Any comments or suggestions would be highly appreciated

Answer 1

Hint

First of all we need $x^2+4y\ge0$

As $\sqrt{x^2+4y}\ge0$

$\sqrt{x^2+4y}>x$ if $x<0$

If $x\ge0,$ we have $$x^2+4y>x^2\iff y>0$$

Answer 2

Claim:$\;$For $x,y\in\mathbb{R}$, the system of inequalities \begin{align*} x-\sqrt{x^2+4y} & < 0\\[4pt] x+\sqrt{x^2+4y} & > 0\\[4pt] \end{align*} is satisfied if and only if $y > 0$.

Proof:

First suppose the system of inequalities is satisfied.$\;$Then \begin{align*} &\left(x-\sqrt{x^2+4y}\right)\left(x-\sqrt{x^2+4y}\right) < 0\\[4pt] \implies\;&x^2-(x^2+4y) < 0\\[4pt] \implies\;&-4y < 0\\[4pt] \implies\;&y > 0\\[4pt] \end{align*} Conversely, suppose $y > 0$.$\;$Then $$x-\sqrt{x^2+4y} < x-\sqrt{x^2} = x-|x| \le 0$$ and also $$x+\sqrt{x^2+4y} > x+\sqrt{x^2} = x+|x| \ge 0$$ so the system of inequalities is satisfied.

Answer 3

1) $x<\sqrt{x^2+4y}$. Two regions to examine:

For $x\ge 0 \implies x^2<{x^2+4y} \implies y>0$

For $x<0 \implies \sqrt{x^2+4y}\ge 0>x\implies y\ge -\frac14 x^2$.

2) $\sqrt{x^2+4y}>-x$. Similarly

For $x>0 \implies \sqrt{x^2+4y}\ge 0>-x\implies y\ge -\frac14 x^2$

For $x\le0 \implies {x^2+4y} > x^2 \implies y>0$.

(1) and (2) together yield $y>0$.