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. 
 A: The reason you're confused is that you should be using the reverse triangle inequality here. What we should have is the following;  $$|z^2+1|^2 \ge ||z|^2-1|^2 = (|z|^2-1)^2$$
Where the last equality is because we will consider the semicircular contour letting $R=|z| \to \infty$ (so certainly bigger than $1$).  
Then we get the result;  $$\left\vert \int_{C_2} \frac{z^2}{(z^2+1)^2} dz \right\vert \le \int_{C_2} \left\vert\frac{z^2}{(z^2+1)^2}\right\vert dz \le \int_0^{\pi} \frac{R^2}{(R^4-2R^2+1)} Rd\theta = \pi\cdot R \cdot \frac{R^2}{(R^4-2R^2+1)} $$.
This tends to zero as $R$ tends to infinity. So the integral around the semi circle $C_2$ disappears.  
What this means, is that when considering our integral around whole contour $C$ and letting $R \to \infty $ the result we get ( which is constant from Cauchy's residue theorem) is simply the contribution from the integral over the real line. Then noting that the integrand is even we halve the result to get to integral between $0$ and $\infty$  
I hope this helps clear up what is going on, and why we do it!!
A: 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 
$$ (\pi R)\max_{z \in C_2} \frac{|z|^2}{|z^2+1|^2} \leq \frac{\pi R^3}{(R^2-1)^2} \rightarrow 0 $$
