= Solution
{c}
Let $\varepsilon>0$. By the definition of <Cauchy sequence>, choose an <integer> $K\geq1$ such that $|x_j-x_k|<\varepsilon/2$ whenever $j,k\geq K$. Apply the given <inequality> with $N=K+1$. It supplies $n>K+1$ for which
$$
0<x_n\leq\frac12\min_{k<K+1}x_k\leq\frac{x_K}{2}.
$$
Both $n$ and $K$ are in the <Cauchy sequence> estimate, so
$$
\frac{x_K}{2}\leq x_K-x_n<\frac\varepsilon2,
\qquad\text{hence}\qquad x_n<\frac\varepsilon2.
$$
For every $m\geq K$ the <triangle inequality> gives
$$
0<x_m\leq|x_m-x_n|+x_n<\varepsilon.
$$
This is the definition of <convergence> to zero: $\boxed{x_m\to0}$. 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 $N=K+1\geq2$ also avoids the empty initial <set> in a literal reading of the condition for very small $N$.
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