Showing that $\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4}$

Question

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.

Answer 1

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.

Answer 2

By the reverse triangle inequality it holds that $|z^4+1| = |1-(-z^4)| \geq 1-|z|^4$ and therefore $\frac{1}{|z^4+1|} \leq \frac{1}{1-|z|^4} = \frac{1}{1-r^4}$

Answer 3

As Lord Shark (indirectly) mentioned, the exponent does not play a role here, as $|z|^n = |z^n|$.

So, it is enough to show that $1-|w| \leq |1 + w|$ for $|w| < 1$, which is a direct consequence of $||a|-|b||\leq |a-b| \stackrel{b \rightarrow -b}{\Longrightarrow} ||a|-|b||\leq |a+b|$