Help with an $\epsilon$, $\delta$ proof

Question

For all $(x,y) \in \mathbf{R}^{2}$ I am trying to find a $\delta > 0$ such that $$\sqrt{(x_{1} -x_{2})^{2} + (y_{1} - y_{2})} < \delta \Rightarrow \left|x_{1} - x_{2}\right| + \left|y_{1}-y_{2}\right| < \epsilon$$ So far I have that \begin{align*} & \left|x_{1} - x_{2}\right| + \left|y_{1}-y_{2}\right| < \epsilon\\ \Rightarrow & \left|x_{1} - x_{2}\right|^{2} + 2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right|+ \left|y_{1} - y_{2}\right|^{2} < \epsilon^{2}\\ \Rightarrow & \left|x_{1} - x_{2}\right|^{2} + \left|y_{1} - y_{2}\right|^{2} < \epsilon^{2} - 2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right|\\ \Rightarrow & \left(x_{1} - x_{2}\right)^{2} + \left(y_{1} - y_{2}\right)^{2} < \epsilon^{2} - 2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right|\\ \Rightarrow & \sqrt{\left(x_{1} - x_{2}\right)^{2} + \left(y_{1} - y_{2}\right)^{2}} < \sqrt{\epsilon^{2} - 2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right|}\\ \end{align*} So if I let $\delta = \sqrt{\epsilon^{2} - 2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right|}$ I can reverse the above to show what I want. The problem is that I need to show that $\epsilon^{2} - 2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right| > 0$ and I am having trouble seeing how to do that or finding a bound for $2 \left|x_{1} - x_{2}\right| \left|y_{1}-y_{2}\right|$.

Answer 1

Notice that $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} < \delta$ implies that both $|x_1-x_2|$ and $|y_1-y_2|$ are less than $\delta$.

This implies that $|x_1-x_2|+|y_1-y_2| < 2\delta$.

So, you can choose $\delta= \epsilon/2$