vanishes. The case of $\alpha=0$ was contained in [@Z1]. Here we give a proof for the remaining case $0<\alpha< \frac{1}{n+2}$ as follows. Let $$\label{e223}
        \eta=\log w-\beta\log(-\hat v)-A|D\hat v|^2,$$ where $w=\hat d^{\alpha-1}$, and $\beta, A$ are positive constants to be determined later. Since $\eta\to+\infty$ on $\partial\hat\Omega_{s,h}$, it attains a minimum at some point $z_0\in\hat\Omega_{s,h}$. At $z_0$, we then have $$\begin{aligned}
        0\!\!&=&\!\!\eta_i=\frac{w_i}{w}-\frac{\eta\hat v_i}{\hat v}-2A\hat v_k\hat v_{ki}, \label{e224}\\
        0\!\!&\leq&\!\![\eta_{ij}]=\left[\frac{w_{ij}}{w}-\frac{w_iw_j}{w^2}-\frac{\beta\hat v_{ij}}{\hat v}+\frac{\beta\hat v_i\hat v_j}{\hat v^2}-2A\hat v_{ki}\hat v_{kj}-2A\hat v_k\hat v_{kij}\right] \label{e225}
    \end{aligned}$$ as a matrix. Since $w=[{\mbox{det}}\,D^2\hat v]^{\alpha-1}$, we have $$\label{e226}
        \hat v^{ij}\hat v_{kij}=(\log{\mbox{det}}\,D^2\hat v)_k=\frac{1}{\alpha-1}\frac{w_k}{w},$$ where $(\hat v^{ij})=\hat d^{-1}(V^{ij})$ is the inverse of $D^2\hat v$. We may assume that $\hat d>1$, otherwise the proof is done. Hence, $$\label{e227}
    \frac{\hat v^{ij}w_{ij}}{w} = \frac{\hat g+\mathcal{\hat F}_t}{\hat d^\alpha} \leq -2\lambda\frac{w_1}{w}(\hat v_1+s)^{-1}-(1-\alpha)(n+2)\lambda\frac{\hat v_{11}}{(\hat v_1+s)^2}+\frac{\sup\hat F_t}{\alpha(\hat v_1+s)^\lambda}.$$ Therefore, we obtain $$\begin{aligned}
\label{e228}
   0 \!\!&\leq&\!\! \hat v^{ij}\eta_{ij} \nonumber\\
     &\leq&\!\! \frac{\sup\hat F_t}{\alpha(\hat v_1+s)^\lambda}+\frac{\lambda(\alpha-1)(n+2)\hat v_{11}}{(\hat v_1+s)^2}-4A\lambda\sum_{k=1}^n\frac{\hat v_{1k}\hat v_k}{\hat v_1+s}-2\lambda\beta\frac{\hat v_1}{(\hat v_1+s)\hat v} \nonumber\\
     && -\frac{\beta n}{\hat v}-\left(2A\bigtriangleup\hat v-\frac{4A^2\alpha}{1-\alpha}\hat v_{ij}\hat v_i\hat v_j\right)-\left(4A\beta-\frac{2A\beta}{1-\alpha}\right)\frac{|D\hat v|^2}{\hat v}-(\beta^2-\beta)\frac{\hat v^{ij}\hat v_i\hat v_j}{\hat v^2} \\
     &\leq&\!\! \frac{\sup\hat F_t}{\alpha(\hat v_1+s)^\lambda}-2\lambda\beta\frac{\hat v_1}{(\hat v_1+s)\hat v}-\frac{\beta n}{\hat v}-\frac{A}{2}\bigtriangleup\hat v-\left(4A\beta-\frac{2A\beta}{1-\alpha}\right)\frac{|D\hat v|^2}{\hat v}, \nonumber
  \end{aligned}$$ with the choice of $\beta>1$ and $A$ small enough such that $$\label{e229}
   \frac{A}{2}\bigtriangleup\hat v \geq \frac{4A^2\alpha}{1-\alpha}\hat v_{ij}\hat v_i\hat v_j+CA^2\hat v_{11},$$ where 