<s>[INST] Least value of $\left|2017 z+\frac{1}{2018z}\right|$ for complex $z$ with $|z|\geq 2019$ Let $z=x+iy$ be a complex number with $|z|\geq 2019$. What is the least value of $$\left|2017 z+\frac{1}{2018z}\right| \, ?$$
What I tried: Put $z=re^{i\alpha}$. Then $z^{-1}=\frac{1}{r}e^{-i\alpha}$ and
$$
\left|2017 re^{i\alpha}+\frac{1}{2018r}e^{-i\alpha}\right| \\
= \left |\left(2017r+\frac{1}{2018r}\right)\cos \alpha+\left(2017r-\frac{1}{2018r}\right)\sin \alpha\right| \\
=\sqrt{\left|\left(2017r+\frac{1}{2018r}\right)^2\cos^2\alpha+\left(2017r-\frac{1}{2018r}\right)^2\sin^2\alpha\right|}
$$
 [/INST] From the triangle inequality
$$
|z+w| \ge |z| - |w|
$$
(with equality if $\overline z w$ is zero or a negative real number) we get
$$
 \left|2017 z+\frac{1}{2018z}\right| \ge 2017 |z|-\frac{1}{2018|z|} 
$$
with equality if $z^2 \le 0$, that is for purely imaginary $z$.
The right-hand side is increasing in $|z|$, so that the minimal value is obtained for the minimal $|z| = 2019$:
$$
\left|2017 z+\frac{1}{2018z}\right| \ge 2017 \cdot 2019 -\frac{1}{2018 \cdot 2019} \, .
$$
Equality holds for $z = \pm 2019 i$.

You can also continue with your approach (using polar coordinates) and substitute $\sin^2 \alpha  = 1 - \cos^2 \alpha$:
$$
\sqrt{\left(2017r+\frac{1}{2018r}\right)^2\cos^2\alpha+\left(2017r-\frac{1}{2018r}\right)^2\sin^2\alpha}
= \sqrt{ 4 \frac{2017}{2018}\cos^2\alpha+\left(2017r-\frac{1}{2018r}\right)^2} 
\ge 2017r-\frac{1}{2018r}
$$
where the minimum is attained if $\cos^2 \alpha = 0$. The rest is as above.
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