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 $\frac{1}{f(z)}=g(z)$.