Existence of a Holomorphic Function Does there exist a function $f(z)$ holomorphic in $\mathbb{C}\backslash\{0\}$, such that
$$\left|f(z)\right|\geq\frac{1}{\sqrt{\left|z\right|}}$$
for all $z\in\mathbb{C}\backslash\{0\}?$
I'm not really sure on how to proceed or which particular theorems I should look at.
 A: I will flesh out my comment.
Let $g(z)=\frac{1}{f(z)}$. Since $|f(z)|\ge\sqrt{|z|}$, $f(z)\neq 0$ on $\mathbb{C}\backslash \{0\}$. Thus, $g(z)$ is holomorphic on $\mathbb{C}\backslash \{0\}$. Furthermore, $|g(z)|=\left|\frac{1}{f(z)}\right|\le\sqrt{|z|}$, so $\lim_{z\to 0}\;g(z)=0$. Therefore, $g(z)$ has a removable singularity at $0$, and so $g(z)$ is entire with $g(0)=0$.
By Cauchy's Integral Formula,
$$
g'(z)=\frac{1}{2\pi i}\int_\gamma \frac{g(w)\;\mathrm{d}w}{(w-z)^2}
$$
Where $\gamma$ is any curve circling $z$ once counterclockwise.  Let $\gamma$ be a circle of radius $R+|z|$ centered at the origin.  Then
$$
|g'(z)|\le\frac{1}{2\pi}\frac{\sqrt{R+|z|}\;2\pi(R+|z|)}{R^2}
$$
Since $R$ is arbitrary, we get that $g'(z)=0$ for all z.  Since $g(0)=0$, we get that $g(z)=0$ for all $z\in\mathbb{C}$. Thus, there can be no $f$ so that $\frac{1}{f(z)}=g(z)$.
A: It has gotten to the point where the main ideas for a solution have already appeared in the comments, so I figured an answer collecting some of these might as well be posted.
Suppose such $f$ exists.  Define $h(z)=1/f(z)$ for $z\neq 0$, and $h(0)=0$.  As Pierre-Yves Gaillard commented, $h$ has the form $h(z)=zg(z)$ for some entire $g$.  Rearranging the original inequality in terms of $g$ shows that $g(z)\to 0$ as $z\to \infty$, and I strongly suspect that you have seen a theorem that will tell you what possible entire functions go to $0$ at infinity.
