Find all entire functions $f$ such that for all $z\in \mathbb{C}$, $|f(z)|\ge \frac{1}{|z|+1}$ Find all entire functions $f$ such that for all $z\in \mathbb{C}$, $|f(z)|\ge \frac{1}{|z|+1}$
This is one of the past qualifying exams that I was working on and I think that I have to find the function that involved with $f$ that is bounded and use Louiville's theorem to say that the function that is found is constant and conclude something about $f$. I can only think of using $1/f$ so that $\frac{1}{|f(z)|} \le |z|+1$ but $|z|+1$ is not really bounded so I would like to ask you for some hint or idea. 
Any hint/ idea would be appreciated. 
Thank you in advance. 
 A: Hint: As you suggest, you can consider $g = 1/f$, which is itself an entire function (why?). Use the estimate $|g(z)|\leq 1 + |z|$ and the Cauchy estimates to show that $g$ is a (nonvanishing) polynomial of degree $\leq 1$. What can you conclude?
A: Suppose $f$ is not constant. As an entire non-constant function it must have some sort of singularity at infinity. It cannot be a pole (because then it would have a zero somewhere, cf the winding-number proof of the FTA), so it must be an essential singularity. Then $zf(z)$ also has an essential singularity at infinity. But $|zf(z)|\ge \frac{|z|}{|z|+1}$, which goes towards $1$, which contradicts Big Picard for $zf(z)$.
A: Here is an alternate proof that does not use Picard's theorem. Clearly $|f(z)|\ge1 $$,\forall z \in \mathbb{C}$ So $f(\mathbb{C})\cap B(0,1)=\emptyset$. So $f$ does not vanish on $\mathbb{C}$! Define $h(z)=\frac{1}{f(z)},z\in \mathbb{C}.$ Then $h$ is entire and $|h(z)|=\frac{1}{|f(z)|}\le1 $. So $h$ is bounded and entire. So, by Louville's thoerem $h$ is constant. Therefore, $f$ has to be constant! 
