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) $$ \begin{equation} \tag{2} \begin{aligned} 0=-\int_{\partial B_1} \int_0^1|\frac{\partial u}{\partial r}|^2 r^{n-1}+\int_{\partial B_1}u \frac{\partial u}{\partial \nu}-\int_0^1r^{n-3}\int_{\partial B_1}|\partial_{\theta}u|^2(r\theta)\\ =-\int_{B_1} |\frac{\partial u}{\partial r}|^2 +\int_{\partial B_1}u \frac{\partial u}{\partial \nu}-\int_{ B_1}|\partial_{\tau}u|^2\\ =-\int_{B_1}|Du|^2+\int_{\partial {B_1}}u\frac{\partial u}{\partial \nu}. \end{aligned} \end{equation} $$