Suppose $u$ is a harmonic function, $\Delta u=0, u\neq0, \psi=|Du|^{2}u^{k},k=-\frac{2(n-1)}{n-2}$, prove: $\Delta\psi\geq0$, when $n\geq3$.
I have tried to calculate the explicit expression for the Laplacian of $\psi$. Here is the Laplacian I got: $$\Delta\psi=k(k-1)u^{k-2}(\sum\limits_{i}u_{i}^{2})^{2}+4ku^{k-1}\sum\limits_{i,j}u_{ij}u_{i}u_{j}+2u^{k}\sum\limits_{i,j}u_{ij}^{2}$$The $u_{i}$ and $u_{ij}$ here are the first and second oreder partial derivatives. If I denote $\sum\limits_{i}u_{i}^{2}$ as $|\nabla u|^{2}$ and let $H$ be the Hessian of $u$. The equation takes the form: \begin{align} \Delta\psi&=k(k-1)u^{k-2}|\nabla u|^{2}+4ku^{k-1}(\nabla u\cdot H\cdot \nabla u^{T})+2u^{k}|H|^{2}\\ &=u^{k-2}(k(k-1)|\nabla u|^{2}+4ku(\nabla u\cdot H\cdot \nabla u^{T})+2u^{2}|H|^{2})\\ &=u^{k-2}f(u) \end{align}
Here the $|\cdot|$ means Frobenius norm of matrix. And $f(u)$ is a quadratic polynimial of $u$. Then I proceed to calculate the discriminant of $f$ and arrived at following equation: $$\Delta=8k(2k(\nabla u\cdot H\cdot \nabla u^{T})^{2}-(k-1)|H|^{2}|\nabla u|^{4})$$ If $\Delta\leq0$ then it is done, however no clues lead to that. Another observation is $(\nabla u\cdot H\cdot \nabla u^{T})^{2}\leq|H|^{2}|\nabla u|^{4}$, I've tried to use this inequality to do something, but in vain. I am wondering if it is true that $2(\nabla u\cdot H\cdot \nabla u^{T})^{2}\leq|H|^{2}|\nabla u|^{4}$? This inequality will be suffice.
