# Complex inequality in Rudin

I assume I'm going to end up slapping my forehead after I see the answer to this, but here goes: in Baby Rudin, Example 5.18 is a demonstration that L'Hopital's rule doesn't hold for functions from $\mathbb{R}$ to $\mathbb{C}$. I follow the reasoning well enough, except for one silly inequality: he asserts that \begin{align} \left|1 + \left\lbrace 2x - \frac{2i}{x}\right \rbrace e^{i/x^2}\right| \geq \left|2x - \frac{2i}{x}\right|-1 \end{align} when $x \in (0,1)$.

I don't see why this is true. It would be nice to apply the triangle inequality, but it doesn't seem to help: we get \begin{align} 1 + \left|2x - \frac{2i}{x}\right| \geq \left|1 + \left\lbrace 2x - \frac{2i}{x}\right \rbrace e^{i/x^2}\right|, \end{align} which seems tantalizing but useless.

Any help would be appreciated.

• In any metric space we have $d(x,y) \geq |d(x,z) - d(z,y)| \geq d(x,z) - d(z,y)$ for all $x,y,z$ in the metric space. – Benicio Jun 15 '17 at 19:11

How about the reverse triangle inequality: $$\lvert \lvert z_1 \rvert - \lvert z_2 \rvert \rvert \leq \lvert z_1-z_2 \rvert,$$

where $z_1 = \left( 2x - \frac{2i}{x} \right) e^{i/x^2}$ and $z_2 =-1$.

If we continue further in the Reverse triangle inequality

$$\overbrace{\left| \underbrace{2x - \frac{2 i}{x}}_{\text{complex number with the real part 2x}} \right| \underbrace{\left| e^{i/x^2} \right|}_{1}}^{\text{range is (0,2)}} -1$$

because of Theorem 1.33d on page 14

$$|z| \ge |Re(z)|$$ and as $x \in (0,1)$