<s>[INST] Showing that $\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4}$ 
I am trying to show that if $|z|=r<1$, then
  $$\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

I have shown the inequality $$\left|\frac{1}{z^3+1}\right|\leq\frac{1}{1-r^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ holds under the same conditions, but I am having showing the result of $(1)$.
I considered \begin{align}
|z^4+1|&\geq\left||z^4|-|-1|\right| \\
&=\left||z|^4-1\right| \\
&=\left|r^4-1\right|
\end{align}
Now, $r^4-1$ is positive $\forall r<1$\ $\{0\}$ $(r\neq -1)$. Ideally, like in $(2)$, if $r^4-1$ was strictly negative then the result would immediately follow.
A hint would be very helpful.
 [/INST] Recall that
$$\left|\frac{1}{z^4+1}\right|=\frac1{\left|z^4+1\right|}$$
and since
$$\left|z^4+1\right|\ge \left|r^4-1\right|=1-r^4\ge 0$$
the result follows.
</s>