So I'm trying to solve some real valued improper integrals with the residue theorem and I have some questions about the curve-contour in the upper half circle. When I want to show that this integral goes to 0 as the radius R goes to infinity I often see people use the estimation lemma(ML) with the triangle inequality, why is that. If you take a look at the integral below
$$\bigl|\int_c \frac{1}{1+z^2+z^4}dz\bigr|$$
Isn't it enough to take whatever integral that is an upper bound for this, for example just
$$\int_c\frac{1}{z^4}dz$$
Im not sure that I fully understand what happens here, maybe someone wants to clear things up for me?
In my opinion you really need to learn the trick using triangle inequality. It is not too hard though. For example in this situation, you want to bound
$$\left| \frac{1}{1 + z^2 +z^4} \right|$$
by $\left| \frac{1}{z^4}\right|$. Now note that
$$|z^4| = |z^4 + z^2 + 1 - z^2 - 1| \le |z^4 + z^2 + 1 | +|z^2| +1$$
$$\Rightarrow |z|^4 - |z|^2 -1 \le |z^4 + z^2 + 1 | .$$
This is not quite you want: there are some negative term there ($-|z|^2 -1$). To get rid of that, you try to make
$$(*) -|z|^2> - \frac 14 |z|^4,\ \ \ \ -1 > -\frac 14 |z|^4$$
so that you can have the inequality $$ |z|^4 - \frac 14 |z|^4 -\frac 14 |z|^4 \le |z^4 + z^2 + 1 | $$ so that $$\frac 12 |z|^4 \le |z^4 + z^2 + 1 | \Rightarrow \frac{2}{|z|^4} \ge \frac{1}{|z^4 + z^2 + 1 |} . $$
One can see that $(*)$ is satisfied if $R = |z|$ is large enough. (In this case $R > \sqrt 2$ suffices).