Find the least value assumed by the function $w=|z+\frac{1}{z}|$ where $,|z|\ge2$
Attempt
By triangular inequality,$|z+\frac{1}{z}|\ge ||z|-|\frac{1}{z}||$
$|z|\ge2......................(1)$
$|\frac{1}{z}|\le\frac{1}{2}$
$-|\frac{1}{z}|\ge\frac{-1}{2}.........(2)$
$|z|-|\frac{1}{z}|\ge2-\frac{1}{2}$
$|z|-|\frac{1}{z}|\ge\frac{3}{2}$
so $|z+\frac{1}{z}|\ge \frac{3}{2}$
But the answer given in my book is $-\frac{1}{8}$
Let $r = |z| \ge 2$ then
$w=|z+\frac{1}{z}| = |r e^{i \phi}+\frac{1}{r} e^{-i \phi}|= r |1 +\frac{1}{r^2} e^{-2 i \phi}|$.
Now the minimum w.r.t. $\phi$ is obtained for $\phi = \pi/2$ and we get
$w \ge r |1 -\frac{1}{r^2}| = r - 1/r$.
Now for $r \ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.