An inequality for harmonic functions Let $u:\overline{B^n}\to \mathbb{R}$ be harmonic, where $\overline{B^n}\subset\mathbb{R}^n$ is the closed unit ball. I would like to prove that
$$
\int_{B_1} |Du|^2\leq \int_{\partial B_1} |\partial_{\tau}u|^2,
$$
where $\partial_{\tau}u$ is the tangential derivative of $u$. I tried to use both the Gauss-Green theorem and to write the Laplacian in spherical coordinates, but I always get stuck.

Here are some correct but apparently useless computations.
Using the Gauss-Green formula we get
$$
\begin{equation}
0=\int_{B_1} u\Delta u =-\int_{B_1}|Du|^2+\int_{\partial {B_1}}u\frac{\partial u}{\partial \nu}.
\end{equation}
$$
The expression of the Laplacian in spherical coordinates is
$$
\begin{equation}
\begin{aligned}
0=\Delta u=\frac{\partial^2u}{\partial r^2} +\frac{n-1}{r}\frac{\partial u}{\partial r}+ \frac{1}{r^2}\Delta_{\partial B_1} u\\
=\frac{1}{r^{n-1}}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u}{\partial r})+ \frac{1}{r^2}\Delta_{\partial B_1} u,
\end{aligned}
\end{equation}
$$
Integrating this against $u$ and noting that $\partial_{\tau}u=1/r\partial_{\theta}u$ we find the same expression as before (obviously)
$$
\begin{equation}
\tag{2}
\begin{aligned}
0=-\int_{\partial B_1} \int_0^1|\frac{\partial u}{\partial r}|^2 r^{n-1}+\int_{\partial B_1}u \frac{\partial u}{\partial \nu}-\int_0^1r^{n-3}\int_{\partial B_1}|\partial_{\theta}u|^2(r\theta)\\
=-\int_{B_1} |\frac{\partial u}{\partial r}|^2 +\int_{\partial B_1}u \frac{\partial u}{\partial \nu}-\int_{ B_1}|\partial_{\tau}u|^2\\
=-\int_{B_1}|Du|^2+\int_{\partial {B_1}}u\frac{\partial u}{\partial \nu}.
\end{aligned}
\end{equation}
$$
 A: Given an open, bounded subset $\Omega$ of $\mathbb R^n$ with $C^1$ boundary and $u \in C^2(\Omega) \cap C^1(\overline \Omega)$, the Rellich-Pohozaev identity is \begin{align*}\int_\Omega (x \cdot \nabla u) \Delta u \, dx &= \int_{\partial \Omega} (\partial_\nu u)(x\cdot \nabla u) \, d\mathcal H^{n-1}_x - \frac 12 \int_{\partial \Omega} \vert \nabla u \vert^2 (x\cdot \nu) \,d\mathcal H^{n-1}_x  + \frac {n-2}2 \int_\Omega \vert \nabla u \vert^2 \, dx 
\end{align*} where $\nu$ is the outward pointing unit normal, $\partial_\nu u :=  \nabla u \cdot \nu$, and $\mathcal H^{n-1}_x$ is $(n-1)$-Hausdorff measure. For a proof, see Theorem 2.1.8 in Elliptic partial differential equations from an elementary viewpoint by Dipierro and Valdinoci (they prove it with Dirichlet boundary conditions, but the proof is the same).
Suppose that $n \geqslant 3$,  $\Omega = B_1$, and $\Delta u =0$ in $B_1$.  Then $\nu = x$, so it follows that \begin{align*}0&= \int_{\partial B_1} (\partial_\nu u)^2\, d\mathcal H^{n-1}_x - \frac 12 \int_{\partial B_1} \vert \nabla u \vert^2 \,d\mathcal H^{n-1}_x  + \frac {n-2}2 \int_{B_1} \vert \nabla u \vert^2 \, dx .
\end{align*} Moreover, by orthogonality of the tangent space and $\nu$, $\vert \nabla u \vert^2 = \vert \partial_\tau u \vert^2 + (\partial_\nu u)^2$, so \begin{align*}
\int_{B_1} \vert \nabla u \vert^2 \, dx &= \frac 2 {n-2} \bigg (\frac 12 \int_{\partial B_1} \vert \nabla u \vert^2 \,d\mathcal H^{n-1}_x - \int_{\partial B_1} (\partial_\nu u)^2\, d\mathcal H^{n-1}_x\bigg ) \\
&= \frac 2 {n-2} \bigg (\frac 12 \int_{\partial B_1} \vert \partial_\tau u \vert^2 \,d\mathcal H^{n-1}_x - \frac 12\int_{\partial B_1} (\partial_\nu u)^2\, d\mathcal H^{n-1}_x\bigg ) \\
&\leqslant \frac 1 {n-2}  \int_{\partial B_1} \vert \partial_\tau u \vert^2 \,d\mathcal H^{n-1}_x .
\end{align*} In particular, $\frac 1{n-2} \leqslant 1$, which gives the result.
I'm not sure about the case $n=2$, I'd have to think about it a bit more.
