An inequality for harmonic functions In a paper that I am reading the author quotes the following result about harmonic functions. According to him this should be "easy to show" but I don't seem to be able to do so.
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
$$
\begin{equation}
0=\int_{B_1} u\Delta u =-\int_{B_1}|Du|^2+\int_{\partial {B_1}}u\frac{\partial u}{\partial \nu}.
\end{equation}
$$
On the other hand, the expression of the Laplacian in spherical coordinates is
$$
\begin{equation}
\begin{aligned}
0=\Delta u=\frac{\partial^2u}{\partial r^2} +\frac{n-1}{r}\frac{\partial u}{\partial r}+ \frac{1}{r^2}\Delta_{\partial B_1} u\\
=\frac{1}{r^{n-1}}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u}{\partial r})+ \frac{1}{r^2}\Delta_{\partial B_1} u,
\end{aligned}
\end{equation}
$$
and 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: Consider first the $d=2$ case. Then, $u$ is a real part of an analytic function. We can write $$u(z)=\frac12\sum_{n=0}^{\infty}(a_nz^n+\overline{a}_n\overline{z}^n)$$ and $$\partial_\nu u(z)=\frac12\sum_{n=0}^{\infty}(a_nnz^n+\overline{a}_nn\overline{z}^n)$$ for $|z|=1$. When multiplying them out and integrating over the circle, we get terms of the form $c\cdot z^m=c\cdot e^{im\theta}$ for some $m\in\mathbb{Z}$, which integrage to zero unless $m=0$. This yields
$$
\int_{\partial B_1} u\partial_\nu u=\pi\sum_{n=0}^\infty n|a_n|^2\leq \pi\sum_{n=0}^\infty n^2|a_n|^2=\int_{\partial B_1} |\partial_\nu u|^2.
$$
The generalization to higher dimensions is straightforward, for one can still write an arbitrary harmonic function as $f(x)=\sum_{n,j}a_{n,j}r^n Y_{n,j}(x/r)$, where $Y_{n,j}$ are spherical harmonics (hence mutually orthogonal).
Update: I misread the question and the above proof proves a different inequality - with $\partial_\nu$ instead of $\partial_\tau$ in the right-hand side - I'm assuming $\partial_\tau$ stands for the spherical gradient.
However, the same idea works in the case of OP's inequality. Note that the spherical harmonics are an orthonormalgonal basis of $L^2(S_1)$, and also $Y_{n,j}$ is an eigenfunction of the Laplace-Beltrami operation on the sphere with eigenvalue $n(n+d-2)$. In particular, they are orthogonal with respect to the Dirichlet form:
$$
\int_{S_1}\partial_\tau Y_{n,j}\cdot\partial_\tau Y_{\hat{n},\hat{j}}= \int_{S_1}\Delta_{S_1} Y_{n,j} Y_{\hat{n},\hat{j}}=n(n+d-2)\delta_{n,\hat{n}}\delta_{j,\hat{j}}.
$$ This means that if $u(x)=\sum_{n,j}a_{n,j}r^n Y_{n,j}(x/r)$, then, as explained above,
$$
\int_{S_1} u\partial_\nu u=\sum_{n,j} a^2_{n,j}n\leq\sum_{n,j} a^2_{n,j}n(n+d-2)=\int_{S_1} |\partial_\tau u|^2.
$$
