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$\displaystyle \left|z^3 + {1 \over z^3}\right| \le 2$ prove that $\displaystyle \left|z + {1 \over z}\right| \le 2$

$$\left|z^3 + {1 \over z^3}\right| = \left(z^3 + {1 \over z^3}\right)\left(\overline z^3 + {1 \over \overline{z}^3}\right) = \left(z + {1\over z}\right)\left(z^2 - 1 + {1\over z^2} \right)\left(\overline z + {1\over \overline z}\right)\left(\overline z^2 - 1 + {1\over\overline z^2} \right)$$

$$=\left|z + {1\over z}\right|^2\left|z^2 - 1 + {1\over z^2} \right|^2 \le 2$$

$$\therefore \left|z + {1\over z}\right| \le \sqrt{2}$$ where $\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1$.

But I am not able to prove $\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1$, need some help on this.

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    There are some mistakes in your computations, in particular, $w\bar w=|w|^2$, not $|w|$ – 2017-02-15
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    @Did Oh yes. That was typo. – 2017-02-15
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    @Did I was think I have missed something. Now I know what. – 2017-02-15

2 Answers 2

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Let $a = \lvert z + \frac 1z \rvert$. Then $$ a^3 = \left\lvert \left( z + \frac 1z \right)^3 \right\rvert = \left\lvert z^3 + 3z + \frac 3z + \frac{1}{z^3} \right\rvert \le \left\lvert z^3 + \frac{1}{z^3} \right\rvert + 3 \left\lvert z + \frac{1}{z}\right\rvert \le 2 + 3a $$ so that $$ 0 \ge a^3 - 3a - 2 = (a-2)(a+1)^2 $$ and therefore $a \le 2$.

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    I understand your proof but is there a way to vaildate my idea ? – 2017-02-15
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    @A---B: I don't know if you can verify $\left|z^2 - 1 + {1\over z^2} \right| \ge 1$ directly. But if you use $z^2 - 1 + {1\over z^2} = (z + {1\over z})^2 - 3$ then you'll get the same inequality as above. – 2017-02-15
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    I am in awe, 9 upvotes for a non-HNQ answer is quite rare. – 2017-02-15
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It's all about identities.

Note that $\left( z + \frac 1z\right)^3 = z^3 + \frac 1{z^3} + 3\left(z + \frac 1z\right)$.

Apply the triangle inequality: $$ \left|\left( z + \frac 1z\right)^3 \right| \leq \left|z^3 + \frac 1{z^3}\right| + 3\left|\left(z + \frac 1z\right)\right| $$ Using what you know: $$ \left|\left( z + \frac 1z\right) \right|^3 - 3\left|\left(z + \frac 1z\right)\right| \leq \left|z^3 + \frac 1{z^3}\right| \leq 2 $$

Suppose that $x = \left|\left( z + \frac 1z\right) \right|$, then $x^3 - 3x \leq 2$ is true, along with $x \geq 0$.

To solve this, note that $x^3-3x = x(x^2-3)$, which are both increasing functions for $x\geq 0$. So, $x^3-3x$ is also increasing. Hence, we only need to find when $x^3 - 3x=2$, which happens at $x=2$. Hence, we can conclude, by the increasing property, that $0 \leq x \leq 2$ is true, but this is the conclusion of the problem.

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    $x^3-3x$ is *decreasing* for $0 < x < 1$. – 2017-02-15