Question

If $\left|z\right|\ge 3,$ prove that the least value of $|z+\frac{1}{z}|$ is $\frac{8}{3}$.

Answer 1

$|z+\frac{1}{z}|\ge \left|z\right|-\frac{1}{\left|z\right|}$
Now $\left|z\right|\ge 3$ $\therefore \frac{1}{\left|z\right|}\le \frac{1}{3}$ or $-\frac{1}{\left|z\right|}\ge -\frac{1}{3}.$
Adding the two like inequalities
$\left|z\right|-\frac{1}{\left|z\right|}|3-\frac{1}{3}=\frac{8}{3}$
Hence from $\left(1\right)$ and $\left(2\right)$, we get $|z+\frac{1}{z}|\ge \frac{8}{3}$
$\therefore$ Least value is $8/3$.