For surfaces of constant mean curvature in $E^3$, prove that either they are all-umbilic-points surfaces or their fundamental forms can be represented as following:
I $=\lambda(u,v)(dudu+dvdv)$
$(\lambda>0)$
II $=(1+\lambda H)dudu-(1-\lambda H)dvdv$
where $H$ is the constant mean curvature.
We can choose proper coordinates so that the coordinate lines are lines of curvature, which implies $F=M=0$ and $H=\frac{LG+NE}{2EG}$. In this case the Codazzi equations can be simplified to $L_v=HE_v$ and $N_u=HG_u$. I tried to deduce $E=G$ from these facts, which will easily lead to the final conclusion, but got stuck here.
Another possible way is to choose the isothemal coordinates. I haven't put too much thought on that yet.
As I said in the question, by integrating the simplified Codazzi equations we have $L=HE+A(u)\ \ N=HG+B(v).\ $ Put this into the expression of H we gain $$\frac{A(u)}{E}=-\frac{B(v)}{G}\triangleq \delta(u,v) $$ and $$(E\delta)_v=0\ \ (G\delta)_u=0 \Rightarrow (ln\delta)_u=-E_u/E\ \ \ (ln\delta)_v=-G_v/G,$$ which implies $$(E_u/E)_{v}=(G_v/G)_{u}\Rightarrow (ln(E/G))_{uv}=0.$$ Integrate this, then there exists $f(u)$ and $g(v)$ such that $$\frac{E}{f^2(u)}=\frac{G}{g^2(v)}$$Define a coordinate transform by$$\tilde{u}=\int f(u)du\quad \tilde{v}=\int g(v)dv$$Under the new coordinate system, coefficients of fundamental forms are$$\widetilde{E}=\frac{E}{f^2(u)}=\frac{G}{g^2(v)}=\widetilde{G}\triangleq \lambda (\tilde{u},\tilde{v})$$ $$\widetilde{F}=\widetilde{M}=0$$By the same deduction$$\widetilde{L}=\lambda H+\widetilde{A}(\tilde{u})$$ $$\widetilde{N}=\lambda H+\widetilde{B}(\tilde{v})$$We may as well set $\widetilde{A}(\tilde{u})=-\widetilde{B}(\tilde{v})=1,$ then we have
I $=\lambda(\tilde{u},\tilde{v})(d\tilde{u}d\tilde{u}+d\tilde{v}d\tilde{v})$ $(\lambda>0)$
II $=(1+\lambda H)d\tilde{u}d\tilde{u}-(1-\lambda H)d\tilde{v}d\tilde{v}$