I'm trying to solve the following exercise:
Identify all holomorphic functions $f:\mathbb C\backslash\{0\} \to \mathbb C$ with the property $\forall z \in \mathbb C \backslash \{0\} :\lvert f(z) \rvert \geq \left \lvert \frac{1}{z} \right \rvert$.
We know that $\lim_{z\to 0}\lvert f(z)| \geq \lim_{z\to 0} \left \lvert \frac{1}{z}\right \rvert = \infty$, so $z_0 = 0$ must be a pole of $f$. If we have a pole of order 1, then $g(z) := \frac{1}{zf(z)}$ is holomorphic on $\mathbb C \backslash \{0\}$ since $f$ has no zeros. Now $\lvert g(z) \rvert \leq 1$, hence bounded with a removable singularity at $z_0 = 0$ and by Liouville already constant. Thereby $f(z) = \frac{c}{z}$ for some constant c. Is that correct? And how do I deal with the case that the pole is of higher order?
With your approach it seems difficult to rule out the possibility that $f$ is something like $$ f(z) = \frac{e^z}{z^5}+ \frac{1}{z} $$ The problem is that there is no obvious place where the $\ge 1/|z|$ inequality fails for a function like this.
As Daniel Fischer suggested, it is better to consider $g = 1/f$ which is also holomorphic in $\mathbb{C}\setminus \{0\}$ and satisfies $|g(z)|\le |z|$. It's easier to use an upper bound on modulus than a lower one... The singularity of $g$ at $0$ is removable, and the rest can go as in: Suppose $f$ is entire and there exist constants $a, b$ such that $|f(z)| \leq a|z|+b$ for all $z \in \mathbb{C}$. Then $f$ is linear
... or even simpler (suggested by zhw.): since $g(0)=0$, the singularity of $g(z)/z$ is removable, so $g(z)/z$ is a bounded entire function which must be constant by Liouville's theorem.