Part c)

Solution.

Firstly for \(x = -y\), \[\lvert \sqrt{1 + x^2} - \sqrt{1 + y^2}\rvert = 0 \leq \lvert -2y \rvert = \lvert x - y \rvert.\] For \(x \neq -y\), we have \[\begin{align} \lvert \sqrt{1 + x^2} - \sqrt{1 + y^2}\rvert &= \frac{\lvert 1 + x^2 - (1 + y^2) \rvert}{\sqrt{1 + x^2} + \sqrt{1 + y^2}},\tag{*}\\ &= \frac{\lvert x^2 - y^2 \rvert}{\sqrt{1 + x^2} + \sqrt{1 + y^2}},\nonumber\\ &= \frac{\lvert x + y \rvert \lvert x - y \rvert}{\sqrt{1 + x^2} + \sqrt{1 + y^2}}.\nonumber \end{align}\] Now, \[\lvert x \rvert \leq \sqrt{1 + x^2}, \;\;\text{and}\;\; \lvert y \rvert \leq \sqrt{1 + y^2}.\] (This can be seen by squaring both sides of each inequality)

So, \[\begin{align*} \lvert x + y \rvert &\leq \lvert x \rvert + \lvert y \rvert \;\;\; \text{(by the triangle inequality)}\\ &\leq \sqrt{1 + x^2} + \sqrt{1 + y^2},\\ \Leftrightarrow \frac{1}{\lvert x + y \rvert} &\geq \frac{1}{\sqrt{1 + x^2} + \sqrt{1 + y^2}}. \end{align*}\] Therefore, \[\begin{align} \lvert \sqrt{1 + x^2} - \sqrt{1 + y^2}\rvert &= \frac{\lvert x + y \rvert \lvert x - y \rvert}{\sqrt{1 + x^2} + \sqrt{1 + y^2}},\nonumber\\ &\leq \frac{\lvert x - y \rvert\lvert x + y \rvert}{\lvert x + y \rvert},\tag{**}\\ &= \lvert x - y \rvert,\nonumber \end{align}\] as required!

You might have a few questions about this:

Q1) Why is 4c) done in this way?

A1) It’s an alternative way to the one in the model solutions, but I think it’s good because it uses some techniques that are useful for the sequences part of the course (e.g. the triangle inequality and step (*)).

Q2) Where on Earth did the case \(x = -y\) come from?

A2) If you look at (**), this expression doesn’t work if \(x = -y\), so you need to consider this separately. It’s not an obvious case until you actually reach (**), but once you realise it, it’s an easy thing to add to the start of your solution.

Q3) What about (*)? Where does this come from?

A3) Recall for \(a,b \in \mathbb{R}\), \[(a-b)(a+b) = a^2 - b^2.\] Taking \(a = \sqrt{1 + x^2}, \;\; \text{and} \;\; b = \sqrt{1 + y^2},\) we have that \[\sqrt{1 + x^2} - \sqrt{1 + y^2} = \frac{(1+x^2)-(1+y^2)}{\sqrt{1 + x^2} + \sqrt{1 + y^2}}.\]