<s>[INST] Significance of the Triangle Inequality After working through a few problems regarding the Cauchy Integral Formula, I'm still a little confused on the significance of the triangle inequality. Why do we use it and what information does it tell us? 
See the following example below. 

Determine $\int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx$

First, we'll create the contour $$ \begin{cases} C_1: Rt & -1 \le t \le 1 \\ C_2:R e^{i\theta} & \ \ \, 0 \le \theta \le \pi \\ C=C_1+C_2 \end{cases} $$
$$ x^2 +1 =0 \Rightarrow x=\pm i \ \ \  ; \ \ \ \ \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx$$
$$ \Rightarrow \int_C \frac{z^2}{(z+1)^2}dz = \int_C \frac{z^2}{[(z+i)(z-i)]^2} 
 = \int_C \frac{z^2}{(z+i)^2 (z-i)^2} $$
$z=-i$ is outside the contour and therefore irrelevant, so choosing $z=i$ for the integration. 
$$ \Rightarrow z_0=i \ \ \text{and} \ \ f(z)=\frac{z^2}{(z+i)^2} $$
$$ f'(z)=\frac{d}{dz}\left[ \frac{z^2}{(z+i)^2} \right] = \frac{2z}{(z+i)^2}-\frac{2z^2}{(z+i)^3}$$
Therefore, by the Cauchy Integration Formula, 
$$ \int_C \frac{f(z)}{(z-i)^2}dz = 2\pi i f'(i) = 2\pi i \left[ \frac{2i}{(2i)^2}-\frac{2i^2}{(2i)^3} \right] = \pi \left[1 + \frac{2}{4i^2} \right] = \pi - \frac{\pi}{2} = \frac{\pi}{2} $$
The following part is where I'm a little confused. I know the work, but am not quite sure why I'm doing it or what the end result is telling us. So, by the triangle inequality, 
$$ \left| z^2 +1 \right|^2 \le \left( \left|z^2 \right| +1 \right) ^2 = z^4 +2\left| z^2 \right| +1 \Rightarrow \left| \int_C \frac{z^2}{(z+1)^2}dz \right| \le \frac{R^2}{R^4 +2R +1} $$
$$ \Rightarrow \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx \le \lim_{R \rightarrow \infty} \int_C \frac{z^2}{(z+1)^2}dz \\ \le \lim_{R \rightarrow \infty} \frac{R^2}{R^4 +2R +1} = 0 $$
$$ \therefore \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2} \left[ \frac{\pi}{2} \right] = \frac{\pi}{4} $$
Any help in deciphering the part in the work I pointed out would be very helpful, thank you for your time. 
 [/INST] A fix on this argument, motivated by Fightclub's observation, is this: We know that for every $R>1$: 
$$ \underbrace{\int_{C_1} \frac{z^2}{(z^2+1)^2}dz}_{\int_{-R}^R \frac{x^2}{(x^2+1)^2}dx} + \int_{C_2} \frac{z^2}{(z^2+1)^2}dz = \pi/2 $$
Hence: 
$$\lim_{R\rightarrow\infty} \int_{-R}^R \frac{x^2}{(x^2+1)^2}dx + \lim_{R\rightarrow\infty} \int_{C_2} \frac{z^2}{(z^2+1)^2} dz = \pi/2 $$
Thus: 
$$ \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)^2}dx  + \lim_{R\rightarrow\infty} \int_{C_2} \frac{z^2}{(z^2+1)^2} dz  = \pi/2$$
So it suffices to show the second integral has a limit of 0. We get: 
$$ \left|\int_{C_2} \frac{z^2}{(z^2+1)^2}dz\right| \leq (\pi R)\max_{z \in C_2} \left[\frac{|z|^2}{|z^2+1|^2}\right] $$
which follows because $C_2$ is half the perimeter of a circle of radius $R$. 
But on $C_2$ we get $|z|=R$. Also, by the reverse triangle inequality we get: 
$$ |z^2+1| \geq |z^2| - 1 = R^2-1$$ 
So for all $z \in C_2$ we get (for all $R>1$): 
$$ \frac{|z|^2}{|z^2+1|^2} \leq \frac{R^2}{(R^2-1)^2} $$
and indeed 
$$ (\pi R)\max_{z \in C_2} \frac{|z|^2}{|z^2+1|^2} \leq \frac{\pi R^3}{(R^2-1)^2} \rightarrow 0 $$
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