Let $\delta_{ij}= 1$ when $i=j$, otherwise $\delta_{ij}=0$. And $g_{ij}=g_{ji}$. The $ |g_{ij} -\delta_{ij}|\le \epsilon$ means that $\forall v=(v_1,...,v_n)\in \mathbb R^n$ there is $$ \left|\sum_{i,j=1}^n v_i(g_{ij}-\delta_{ij})v_j \right|\le \epsilon\sum_{i=1}^n v_i^2 \tag{1} $$ where $|\cdot|$ of (1) is absolute value (the following symbol $|\cdot|$ also represents absolute value). I am interested in understanding the inequality for $\det(g_{ij})$ when the $\epsilon>0$ is sufficiently small.
For $n=2$, I guess $$ |\det(g_{ij})|\le (1+\epsilon)^2 $$ But I don't know how to prove. I guess the above inequality just by trying some examples.
Besides, for general $n$, I feel there should be $$ |\det(g_{ij})|\le (1+\epsilon)^n $$