Suppose $u\in C^1(\bar{U})\cap C^3(U)$, where $U$ is a bounded simply connected open set. If $\Delta u=0$ and $u(x)\neq0$ for all $x$ , show that $\varphi=\frac{\vert Du\vert^2}{u^{\frac{2(n-1)}{n-2}}}$ satisfies $\Delta\varphi\geqslant0$
I have calculated that $$\Delta\varphi=2u^{-\frac{2(n-3)}{n-2}}\sum_{i,j=1}^n(u_{ij}^2u^2-4\frac{n-1}{n-2}u_iu_ju_{ij}u+\frac{(n-1)(3n-4)}{(n-2)^2}u_i^2u_j^2)$$ But I don't know what to do next to show $\Delta\varphi\geqslant0$. I also considered that $\varphi=\vert D(u^{-\frac{1}{n-2}})\vert^2$, but I don't make any progress. How should I do next? Any answer or counterexample would be appreciated! Thanks for attention.
Well my teacher made a mistake on the problem. We only need to show that $\varphi$ satisfies the maximal principle and we do not need to show that $\Delta\varphi\geqslant0$.
To this end, let us first consider $\phi=\frac{\vert Du\vert^2}{u^{\alpha}}$ for $\alpha\in(0,\frac{2(n-1)}{n-2})$. If $\phi$ attains its maximum at $x_0\in U$, then $D\phi(x_0)=0$. Take a special coordinate such that at $x_0$ we have $Du(x_0)=(\vert Du\vert,0,\dots,0)$. Then we can compute$$D_i\phi(x_0)=2\sum_{j=1}^nD_juD_{ij}uu^{-\alpha}-\alpha\vert Du\vert^2D_iuu^{-\alpha-1}=2\vert Du\vert D_{i1}uu^{-\alpha}-\alpha\vert Du\vert^3\delta_{i1}u^{-\alpha-1}$$Now there are two cases. If $Du(x_0)=0$, then $\phi(x_0)=0$ and since $\phi(x_0)=\max_U\phi$, we have $\phi=0$ in $U$ hence $\phi$ satisfies the maximal principle. Otherwise, we have $2D_{i1}u=\alpha\vert Du\vert^2u^{-1}\delta_{i1}$ at $x_0$. Next we have \begin{equation} \Delta\phi=2\vert D^2u\vert^2u^{-\alpha}+2D\vert Du\vert^2\cdot Du^{-\alpha}+\vert Du\vert^2(\alpha(\alpha+1)\vert Du\vert^2u^{-\alpha-2} \end{equation}At $x_0$, $$\vert D^2u\vert=(D_{11}u)^2+\sum_{i,j=2}^n(D_{ij}u)^2\geqslant(D_{11}u)^2+\sum_{i=2}^n(D_{ii}u)^2\\\geqslant(D_{11}u)^2+\frac{1}{n-1}(\sum_{i=2}^nD_{ii}u)^2=\frac{n\alpha^2}{4(n-1)}\vert Du\vert^4u^{-2}$$Also, $D\vert Du\vert^2\cdot Du^{-\alpha}=-\alpha^2\vert Du\vert^2u^{-\alpha-2}$ at $x_0$. Hence $$\Delta\phi(x_0)\geqslant(\frac{n\alpha^2}{2(n-1)}-2\alpha^2+\alpha(\alpha+1))\vert Du\vert^2 u^{-\alpha-2}>0$$since $(\frac{n\alpha^2}{2(n-1)}-2\alpha^2+\alpha(\alpha+1))=\alpha^2(\frac{1}{\alpha}-\frac{n-2}{2(n-1)})>0$. However, this contradicts the assumption that $\phi(x_0)=\max_U\phi$. As a result, we get $$\max_U\phi=\max_{\partial U}\phi$$Just let $\alpha\rightarrow \frac{2(n-1)}{n-2}$.