If $h:\mathbb{R}^n \to\mathbb{R}$ is harmonic and $|h(x)|\leq C\sqrt{|x|}$ for all $x\in\mathbb{R}^n$ then $h$ is constant. 
If $h:\mathbb{R}^n \to\mathbb{R}$ is harmonic and $|h(x)|\leq C\sqrt{|x|}$ for all $x\in\mathbb{R}^n$ then $h$ is constant.

Since $|h(x)| \leq C\sqrt{|x|}$ we have that,
$$-C|x|^{1/2} \leq h(x) \leq C|x|^{1/2}$$
Then,
$$-\dfrac{Cx}{2}|x|^{-3/2} \leq \nabla h \leq \dfrac{Cx}{2}|x|^{-3/2}$$
Now I want to show the left and right hand side are $0$ using the fact that $\Delta h = 0$ for all $x\in\mathbb{R}^n$, which would give me my result. Is there a better way to go about this?
 A: You can do this with the mean value theorem if you know the following lemma. 
Lemma. let $x_0, y_0 \in \mathbb R^n$. Then there is $C > 0$ such that $$\text{Vol}(B(x_0, R) \triangle B(y_0, R)) \le C \cdot R^{n-1}$$ for all $R$ sufficiently large. 
Here $\triangle$ denotes the symmetric difference (the elements that are in one of the sets but not both); i.e. $$A \triangle B = (A \setminus B) \cup (B \setminus A)$$ and $B(x_0, R)$ denotes the ball of radius $R$ centered at $x_0$.  I can't find a totally elementary proof of this lemma but it is a special case of a much richer theorem proven here: https://arxiv.org/pdf/1010.2446.pdf. 
We proceed assuming the lemma. Fix $x_0, y_0 \in \mathbb R^n$ and let $R >0$ be large. By the mean value theorem, we have \begin{align*}
\lvert h(x_0) - h(y_0) \rvert &= \frac{\alpha_n}{R^n}\left \lvert \int_{B(x_0, R)} h(x) dx - \int_{B(y_0,R)} h(x) dx \right \rvert \\& \le \frac{\alpha_n}{R^n} \int_{B(x_0,R) \triangle B(y_0, R)} \lvert h(x) \rvert dx. 
\end{align*} However, using the bound on $h$, the largest that $\lvert h(x) \rvert$ can be on $B(x_0,R) \triangle B(y_0,R)$ is some constant times $(R + \max\{\lvert x_0 \rvert, \lvert y_0 \rvert \})^{1/2}$ which is less than $(2R)^{1/2}$ when $R$ is large enough. Thus $$\lvert h(x_0) - h(y_0) \rvert \le C\frac{R^{1/2}}{R^n} \text{Vol}(B(x_0, R) \triangle B(y_0, R)) \le C \frac{R^{n-1/2}}{R^n} = \frac{C}{R^{1/2}}$$ (where I have grouped many constants together into $C$). Since this holds for all $R$ sufficiently large, we send $R \to \infty$ to see that $h(x_0) = h(y_0)$ and since $x_0, y_0$ were arbitrary this shows that $h$ is constant. Of course, if the bound holds for all $x \in \mathbb R^n$, then setting $x =0$ shows that $h = 0$ identically. 
EDIT: The proof of the lemma is actually very easy. As @orangeskid points out in the comments, for any $z \in \mathbb R^n$ we see that $B(x_0 + z, R) \setminus B(x_0,R)$ is a subset of $B(x_0, R + \lvert z \rvert) \setminus B(x_0, R)$ [this realization is trivial upon drawing a picture]. Using this, we can bound the volume of the symmetric difference: \begin{align*}\text{Vol}(B(x_0, R) \triangle B(y_0, R)) &\le B(x_0, R + \lvert x_0 - y_0\rvert) + B(y_0, R + \lvert x_0 - y_0\rvert) \\&= 2\alpha_n ((R+\lvert x_0 - y_0 \rvert)^n - R^n) \lesssim R^{n-1} \end{align*} by the binomial theorem.
