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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.

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    $\begingroup$ This is the same as saying, that if $|w|<1$, then $|1+w|\ge 1-|w|$. $\endgroup$ – Angina Seng Aug 3 '18 at 7:54
  • $\begingroup$ By the reverse triangle inequality? $\endgroup$ – user557493 Aug 3 '18 at 7:56
  • $\begingroup$ $|z^4+1|\geq 1- |z|^4=1-r^4>0.$ $\endgroup$ – Riemann Aug 3 '18 at 7:56
  • $\begingroup$ I'm still unclear. It looks like you're using the reverse triangle inequality, which I thought was defined as $\left|z_1-z_2\right|\geq\left||z_1|-|z_2|\right|$ for every complex number $z_1, z_2$. It looks as though you have assumed that $|z_1|-|z_2|$, or in this case, $1-r^4$ is positive. I don't think this is the case. $\endgroup$ – user557493 Aug 3 '18 at 8:03
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    $\begingroup$ math.stackexchange.com/questions/2870628/… $\endgroup$ – lab bhattacharjee Aug 3 '18 at 8:13
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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.

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  • $\begingroup$ How is $1-r^4\geq 0$ when $r$ is defined as $r<1$? $\endgroup$ – user557493 Aug 3 '18 at 8:04
  • $\begingroup$ r is no negative that is $0\le r<1$ $\endgroup$ – user Aug 3 '18 at 8:08
  • $\begingroup$ This is the part I don't understand. I'm unsure of why $r$ cannot be negative? Now that you say it, I agree that the inequality $(1)$ doesn't hold if $r<0$. How did you see this immediately? I overlooked this completely and assumed that $r\in(-\infty,0)$ as $r<1$. $\endgroup$ – user557493 Aug 3 '18 at 8:14
  • $\begingroup$ recall that $ r=|z|\ge 0$ $\endgroup$ – user Aug 3 '18 at 8:15
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    $\begingroup$ Now it’s clear! Well done. Bye $\endgroup$ – user Aug 3 '18 at 8:18
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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 $||a|-|b||\leq |a-b| \stackrel{b \rightarrow -b}{\Longrightarrow} ||a|-|b||\leq |a+b|$

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

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  • $\begingroup$ Note that the reverse triangle inequality says that $|z-w| \geq | |z| - |w| |$ but this implies that $|z-w| \geq |z| - |w| $ because if $ |z| - |w| < 0$ then the equality holds trivially since $|z-w| \geq 0$. $\endgroup$ – Andreas Dahlberg Aug 3 '18 at 8:30

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