 Solution

Let $\epsilon >0$. By the definition of Cauchy sequence, choose an integer $K\ge 1$ such that $|x_j-x_k|<\epsilon /2$ whenever $j,k\ge K$. Apply the given inequality with $N=K+1$. It supplies $n>K+1$ for whichBoth $n$ and $K$ are in the Cauchy sequence estimate, soFor every $m\ge K$ the triangle inequality givesThis is the definition of convergence to zero: $\boxed{x_m\to 0}$. The proof uses only the Cauchy sequence estimate, positivity, and the stipulated inequality; it does not assume the general completeness of the real numbers theorem for Cauchy sequences. Taking $N=K+1\ge 2$ also avoids the empty initial set in a literal reading of the condition for very small $N$.