if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant?

Since $f\in\mathcal{O}(\mathbb{C})$, then $$ f(z)=\sum\limits_{n=0}^\infty c_n z^n $$ for all $z\in \mathbb{C}$. Moreover, for all $R>0$ we have integral representation for coefficients $$ c_n=\frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz $$ Then, we get an estiamtion $$ |c_n|\leq\frac{1}{2\pi} \oint\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz|\leq \frac{1}{2\pi}\oint\limits_{\partial B(0,R)}\frac{1+|z|^{1/2}}{|z|^{n+1}}|dz|= \frac{1}{2\pi}\frac{2\pi R(1+R^{1/2})}{R^{n+1}} $$ Hence for $n\in\mathbb{N}$ we obtain $$ |c_n|\leq\frac{1}{2\pi}\lim\limits_{R\to+\infty}\frac{2\pi R(1+R^{1/2})}{R^{n+1}}=0 $$ which implies $c_n=0$. Finally we get $$ f(z)=c_0+\sum\limits_{n=1}^\infty c_n z^n=c_0=\mathrm{const} $$ Here you can find generalized version of this answer

A slightly different way: $|z f(1/z)| \leq |z| + |z|^{1/2}$ for $z \neq 0$ so $z f(1/z)$ extends to an entire function $\sum_{k \geq 1} a_k z^k$ by Riemann's extension theorem. Then $f(z) = \sum_{k \geq 1} a_kz^{1-k}$. This implies that all coefficients $a_k$ vanish except possibly $a_1$.