Yes Suppose $e_1,\dots, e_n$ are length of the edges in non-increasing order. COnsider the 2 faces that share the edge $e_1$. So $e_2+\dots e_n>2e_1$. (Write the triangle inequlaity for the 2 faces) Now beacuse $e_i,e_{i+1},e_{i+2}$ can not be edges of triangle so $e_{i+1}+e_{i+2}\leq e_i$ and $2(e_2+\dots+e_n)=e_2+(e_2+e_3)+\dots+(e_{n-1}+e_n)+e_n$ and $e_2+(e_2+e_3)+\dots+(e_{n-1}+e_n)+e_n\leq e_2+e_1+e_2+\dots +e_{n-2}+e_n$ Therefore $e_2+e_3+\dots+e_n\leq e_1+e_2-e_{n-1}<e_1+e_1+0=2e_1$. Contradiction prove the problem and there exist 3 edges that can be sides of triangle.