# Proving Absolute Value Inequality

I had posted a portion of this earlier asking about how to interpret min(). I received some excellent answers, however, I have run into problems and feel stuck. I am posting the question in its entirety.

Let $x, x_{0},y,y_{0}$ be real numbers and $\varepsilon$ be a positive real number. If we have

$$|x-x_{0}|<\min \left(\frac{ \varepsilon }{2(|y_{0}|+1)}, 1\right) \text{ and } |y-y_{0}|< \frac{ \varepsilon }{2(|x_{0}|+1)},$$ then prove that $|xy-x_{0}y_{0}|<\varepsilon$.

A hint is provided:

Write $xy-x_{0}y_{0}$ in terms of $x-x_{0}$ and $y-y_{0}$ and use the triangle inequality twice.

I've been rearranging and writing out what I know etc. in an attempt to find a solution:

$$|x-x_{0}|(2|y_{0}|)+2|x-x_{0}|<\varepsilon$$

$$|y-y_{0}|(2|x_{0}|)+2|y-y_{0}|<\varepsilon$$

$$|x-x_{0}|< 1$$

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HINT: $xy - x_0y_0 = xy - xy_0 + xy_0 - x_0 y_0$. –  JavaMan Sep 21 '11 at 21:02
$|x(y-y_{0})-y_{0}(x-x_{0})|<\varepsilon$ –  Malthus Sep 21 '11 at 21:52
I've got $|x(y-y_{0})-y_{0}(x-x_{0})|\leq |x(y-y_{0})|+|y_{0}(x-x_{0})|$ but I cant go further and relate it back to the inequalities I'm provided with. –  Malthus Sep 21 '11 at 22:24
Also $|x(y-y_{0})|+|y_{0}(x-x_{0})|=|x||(y-y_{0})|+|y_{0}||(x-x_{0})|$ –  Malthus Sep 21 '11 at 22:28

$$\begin{eqnarray}|xy-x_0y_0|&\leq&|xy-x_0y+x_0y-x_0y_0|\\ &\leq&|y||x-x_0|+|x_0||y-y_0|\\ &\leq&(|y_0|+|y-y_0|)|x-x_0|+|x_0||y-y_0|\\ &\leq& |y_0||x-x_0|+|y-y_0|(|x_0|+|x-x_0|)\\ &<&|y_0|\frac{\varepsilon}{2(|y_0|+1)}+\frac{\varepsilon}{2(|x_0|+1)}(|x_0|+1)\\ &<&2\frac{\varepsilon}{2}=\varepsilon. \end{eqnarray}$$
And between the passages we have used the fact that the condition $$|x-x_0|< \min\left(\frac{\varepsilon}{2(|y_0|+1)},1\right),$$ implies both $$|x-x_0|<\frac{\varepsilon}{2(|y_0|+1)}\quad\text{and}\quad |x-x_0|<1.$$