Showing that $\left|\frac{z}{z^3+1}\right|\leq\frac{R}{R^3-1}$

I am trying to show that if $|z|=R>1$, then $$\left|\frac{z}{z^3+1}\right|\leq\frac{R}{R^3-1}$$

I've been playing around with the above inequality and this is where I am at.

\begin{align} \text{Consider} \ \left|z^3+1\right|&\leq\left|z^3\right|+1 \ \ \ \ \ \text{(by the triangle inequality)}\\ &=\left|z\right|^3+1 \\ &=R^3+1 \\ \\ \Rightarrow \left|z^3+1\right|&\leq R^3+1 \\ \frac{1}{\left|z^3+1\right|}&\geq\frac{1}{R^3+1} \\ \frac{|z|}{\left|z^3+1\right|}&\geq\frac{|z|}{R^3+1} \ \ \ \ \ \ \text{(inequality unchanged, |z|>1)} \\ \left|\frac{z}{z^3+1}\right|&\geq\frac{R}{R^3+1} \\ \end{align}

I'm unsure of how to yield the desired inequality. Have I made a mistake somewhere?

EDIT

If I used the inequality $$|z_1+z_2|\geq |z_1|-|z_2|$$ I believe this will work. My question is, does this inequality hold for real numbers as well?

• The triangle $|a|-|b| \le |a+b| \le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well. – Doug M Aug 3 '18 at 1:21

Since $|z|=R$, your inequality reduces to $$\tag1 \frac1{|z^3+1|}\leq\frac1{R^3-1},$$ which in turn is the same as $$\tag2 R^3-1\leq |z^3+1|.$$ And this is just the reverse triangle inequality: $$|z^3+1|\geq |z^3|-1=|z|^3-1=R^3-1.$$

As for your edited question: $$|z_1|=|z_1-z_2+z_2|\leq|z_1-z_2|+|z_2|.$$ This gives $$|z_1|-|z_2|\leq |z_1-z_2|.$$ Since the roles are interchangeable, we get $$|\,|z_1|-|z_2|\,|\leq |z_1-z_2|.$$

• I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|\leq \left||z|-|z_2|\right|$? How did you obtain $|z^3+1|\geq|z^3|-1$? – user557493 Aug 3 '18 at 1:20
• $z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$. – Martin Argerami Aug 3 '18 at 1:26
• I think I typed this wrong, I meant $\left||z_1|-|z_2|\right|\leq|z_1-z_2|$. Is this correct? – user557493 Aug 3 '18 at 1:29
• Yes, that's exactly what I used and proved in my answer. – Martin Argerami Aug 3 '18 at 1:29
• Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|\geq |b|$, then $|a|\geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$. – Martin Argerami Aug 3 '18 at 1:50

Your inequality is correct, but it's going the wrong way. You want an inequality that ends up with $|z^3+1|$ being greater than something. Try starting with: $$|z^3| = |z^3 +1 + (-1)|\le |z^3+1| + |-1|$$

WLOG $z=R(\cos t+i\sin t)$ where $t$ is real

$$|z^3+1|^2=R^6+1+2R^3\cos3t\ge R^6+1-2R^3= (R^3-1)^2$$

as $\cos3t\ge-1$ and $R^3>1$