Denote the numbers as $x_1<\ldots<x_n$. Let $c=\sqrt[n-2]2$ and suppose the contrary. It follows that $x_{i+1}+x_j>c(x_i+x_j)$ for all $1\le i<n$ and $1\le j\le n$. In particular, $x_{i+1}>cx_i+(c-1)x_n$ for all $1\le i<n$. By induction on $i$, it follows that $x_{i+1}>c^ix_1+(c^i-1)x_n$. Take $i=n-2$ and get $x_{n-1}>x_n$, since $x_1>0$, a contradiction.