Bound $x, y$ if we know bounds of $x+y$ and $xy$

Question

If we know $-c < x+y < c$ and $-c < xy < c$ with $c > 0$ and $x, y \in \mathbb R$, what is a good bound for $x$ and $y$?

Intuitively, $x, y$ are in the region that are intersected by two lines $x+y= \pm c$ and four graphs $y=\pm \frac{c}{x}$. For example, a plot like this (with $c=10$). But I don't know how to analytically argue this?

Answer 1

First, note that $|xy|=|x||y|$ and hence if $|xy|<c$, it follows that $|x|<c/|y|$.

Next, note that $|x+y|\geq||x|-|y||\geq|x|-|y|$ and hence if $|x+y|<c$, then $|x|<|y|+c$.

Put these bounds together and you have $|x|<\min\{c/|y|,|y|+c\}$.

Consider now the curves $y\mapsto c/y$ and $y\mapsto y+c$ defined on $[0,\infty)$. These lines intersect at points $y$ satisfying $$ \frac{c}{y}=y+c. $$ By the quadratic formula, the only such point is $$ y=\frac{\sqrt{c^{2}+4c}-c}{2}. $$ The value of the curves at this point is $$ \frac{2c}{\sqrt{c^{2}+4c}-c}. $$ Since the first curve is decreasing and goes to infinity as $y$ vanishes while the second curve is increasing and finite at zero, it follows that the above is an upper bound of $|x|$.

Since $x$ and $y$ are symmetric, this is also true for $|y|$.

Answer 2

With the information: $-c < x+y < c$ and $-c < xy < c$ with $c > 0$ and $x, y \in \mathbb R$, let us consider the boundaries as the curves:

$xy = c$

$xy = -c$

$x + y = c$

$x + y = -c$

With these curves we see that the largest value for x on the boundary is at an intersection between $x + y = c$ and $x y = -c$.

The x-value is the larger root for the equation $x^2 - c x - c$ :

$x_m = \frac{c+\sqrt{c^2 + 4c}}{2}$

Similarly, the maximum y-value for the boundaries are found for the intersection between the same curves:

$y_m = \frac{c+\sqrt{c^2 + 4c}}{2}$

Noting symetry of the curves, the bounds for x and y are:

$ \frac{-c-\sqrt{c^2 + 4c}}{2} < x < \frac{c+\sqrt{c^2 + 4c}}{2} $,

$ \frac{-c-\sqrt{c^2 + 4c}}{2} < y < \frac{c+\sqrt{c^2 + 4c}}{2} $,