I am trying to complete the following exercise from "Topology Without Tears" by Sidney A. Morris (accessible at http://www.topologywithouttears.net/topbook.pdf). It is Exercise 2.2.1, and reads:
(i) Let $\langle a,b \rangle$ be any point in the disk $D = \{\langle a,b \rangle:x^2 + y^2 < 1\}$. Put $r = \sqrt{a^2 + b^2}$. Let $R_{\langle a,b \rangle}$ be the open rectangle with vertices at the points $\langle a \pm \frac{1-r}{8},b \pm \frac{1-r}{8} \rangle$. Verify that $R_{\langle a,b \rangle} \subset D$.
My Approach:
I am restricting myself to the case where $a>0$ and $b>0$. If I can prove that the top right vertex of the open rectangle lies in $D$, then the entire rectangle will lie in $D$. Thus, I must prove that $(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < 1$. Expanding this, I acquire:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 = r^2 + (a+b)\frac{1-r}{4} + 2(\frac{1-r}{8})^2.$$
From here, I keep hitting dead-ends. Using $a+b<2$ gives the continuation:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < r^2 + \frac{1}{2}(1-r) + 2(\frac{1-r}{8})^2.$$
I multiplied out this expression to acquire:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < \frac{33}{32}r^2 - \frac{9}{16}r + \frac{17}{32}.$$
If I factor $r$ out from the first two terms on the RHS, we acquire:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < r(\frac{33}{32}r - \frac{9}{16}) + \frac{17}{32}.$$
If $\frac{33}{32}r > \frac{9}{16}$, then $r < 1$ implies:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < r(\frac{33}{32}r - \frac{9}{16}) + \frac{17}{32} < \frac{33}{32}r - \frac{9}{16} + \frac{17}{32}$$
$$<\frac{33}{32} - \frac{9}{16} + \frac{17}{32} = 1.$$
But if we don't have $\frac{33}{32}r > \frac{9}{16}$, we can't do this since the expression in the parentheses would be negative, and so removing $r$ would no longer make the expression larger. I'm thinking there must be a simpler way to prove this.
EDIT: Solution Possibly Found With help from the answer I have marked as most helpful below, I believe I have arrived at a relatively complete solution. It is easy to see geometrically that we must have $a+b < \sqrt{2}r$, giving us a stronger bound than $a+b<2$. To see this more rigorously, I reference this other question on MathExchange: Prove that $2|ab| \leq a^2 + b^2$ and $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$.
Proceeding, we have:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 = r^2 + (a+b)\frac{1-r}{4} + 2(\frac{1-r}{8})^2$$
$$< r^2 + \sqrt{2}r\frac{1-r}{4} + 2(\frac{1-r}{8})^2 = (r+\sqrt{2}(\frac{1-r}{8}))^2 = ((1-\frac{\sqrt{2}}{8})r + \frac{\sqrt{2}}{8})^2 < 1^2 = 1$$
Hint: Draw a picture. The line $x+y = \sqrt{2}$ is tangent to the unit circle, so $a + b < \sqrt{2}$. That should be enough stronger than $a+ b < 2$ to finish the proof. Maybe $a+b < \sqrt{2} < 3/2$ will do.