If $u,v$ are Harmonic Conjugates in $\mathbb{C}$ and $u^3-3uv^2\ge 0$, then $u,v$ are Constants.

Question

Let $u,v$ be harmonic in $\mathbb{C}$ and assume that $v$ is the harmonic conjugate of $u$. Assume that $$u^3-3uv^2\ge 0\tag{$1$}$$ in $\mathbb{C}$. Prove that $u,v$ are constants.

I am not sure how to use $(1)$ and the LHS of $(1)$ need not be harmonic. Since $u,v$ are conjugates, $u_x=v_y,u_y=-v_x$. Since $u,v$ are harmonic, $u_{xx}+u_{yy}=v_{xx}+v_{yy}=0$. Moreover, $$u_{xx}=v_{xy}=v_{yx}\\v_{yy}=-u_{yx}=-u_{xy}\\u_{yy}=-v_{xy}=-v_{yx}\\v_{yy}=u_{yx}=u_{xy}$$ I am not sure how to use any of these equations since $(1)$ is an inequality and $(1)$ does not deal with derivatives. How can I show $u,v$ are constant?

Answer 1

Define the analytic function $f=u+iv$. Then $z \mapsto f^3(z)$ is also analytic and its real part is $u^3-3uv^2$, which you insist is nonnegative. By Liouville's theorem the function $f^3$ is constant. By continuity, it must be that $f$ is constant.