How to show $|2z^{3} - 4z - 3- i| \geq \sqrt{10} - \frac{5}{2}$ for $|z| \leq \frac{1}{2}$? Assuming $|z| \leq \frac{1}{2}$, show that$$|2z^{3} - 4z - 3- i| \geq \sqrt{10} - \frac{5}{2}.$$
My steps are as follows:
\begin{align*}
|2z^{3} - 4z - 3- i| &= |2z(z^{2} - 2) - (3 + i)|\\
&\geq \bigl| |2z(z^{2}-2)| - |(3+i)|\bigr|\\
&= \bigl||2z(z^{2} -2)| - \sqrt{10}\bigr|\\
&=\Bigl| 2z \bigl||z^{2}| - 2|\bigr| - \sqrt{10}\Bigr|\\
&=\Bigl|\bigl|2| z ||z|^{2} - 2|\bigr| - \sqrt{10}\Bigr|
\end{align*}
So the minimum value is attained if $|z| = 0$, which gives us
\begin{align*}
& \left|\left|0 - 2\right| - \sqrt{10}\right|\\
&= \sqrt{10} - 2 > \sqrt{10}-\frac{5}{2}.
\end{align*}
Since the smallest possible value is greater than $\sqrt{10}-\frac{5}{2}$, we have the result.
 A: Following up on previous comments:
$$
\begin{align}
|2z^{3} - 4z - 3- i| &\geq \left| \,|2z(z^{2}-2)| - |(3+i)|\,\right| \\
 &= \left|\,\left|2z(z^{2} -2)\right| - \sqrt{10}\,\right| \\
 &= \left|\, 2 \left|z\right| \left|z^{2}-2\right| - \sqrt{10}\right| \\
 &\ge \left|\,2 \left|z\right| \left||z|^{2} - 2|\right| - \sqrt{10}\,\right| \\
 &\color{red}{\ge} \left|\,2 \cdot \frac{1}{2} \,\left|\left(\frac{1}{2}\right)^{2} - 2\right| - \sqrt{10}\,\right| \tag{*}\\
 &= \sqrt{10} - \frac{7}{4} \\
 &\gt \sqrt{10} - \frac{5}{2}
\end{align}
$$
The (*) $\,\color{red}{\ge}\,$ step follows from the function $\,|x(x^2-2)|\,$ being easily verified to be strictly increasing for $\,x \in [0, \frac{1}{2}]\,$, and also lower than $\,\sqrt{10}\,$.

[ EDIT ] As a side note, I still can't quite guess what shortcut would directly (and, presumably, more easily) prove the particular lower bound proposed in the question. The above gives a better lower bound $\,\sqrt{10} - \frac{7}{4}\,$ $\simeq 1.41$ $\gt \sqrt{10} - \frac{5}{2}$ $\simeq 0.66\,$, but that's still far from the best lower bound of $\simeq 1.6 \simeq \frac{\sqrt{41}}{4}$ at $\,x = -\frac{1}{2}\,$ if one were to trust WA's min |2z^3-4z-3-i| where 0 <= |z| <= 1/2.
