Let $\Sigma\subset\mathbb{R}^3$ be a regular surface. Prove that $\Sigma$ is minimal if and only if its coordinate functions are harmonic.
I know that every regular surface admits an isothermal parametrization. Assuming that a parametrization $X:U\subset\mathbb{R}^2\to\Sigma$ is isothermal (i.e., $<X_u,X_u>=<X_v,X_v>$ and $<X_u,X_v>=0$), I know how to prove $X_{uu}+X_{vv}=2<X_u,X_u>^2HN$, where $H$ is the mean curvature and $N=\frac{X_u\wedge X_v}{||X_u\wedge X_v||}$. So obviously $H\equiv 0 \iff X_{uu}+X_{vv}\equiv 0$.
The problem doesn't say anything about $X$ being isothermal, so I'm stuck with the general case. I tried to use brute force, but it was such a ridiculous amount of work that I eventuay gave up.
Is there some clever way around it?
The correct result is "if $X$ is isothermal, then $\Sigma$ is minimal if and only if $X$ is harmonic".
Consider the following counter-example: take $w_1,w_2 \in \Bbb R^3$ linearly independent such that $\|w_1\| = 1$ and $\|w_2\| = 2$, and put $X\colon \Bbb R^2 \to \Sigma \doteq {\rm span}(w_1,w_2)$ given by $X(u,v) = uw_1+vw_2$. Clearly $\Sigma$ is minimal but $\langle X_u,X_u\rangle \neq \langle X_v,X_v\rangle$.
The bottom line is: you already "won" at $H \equiv 0 \iff X_{uu}+X_{vv} = 0$.