= Solution
Two <norms> $\|\cdot\|_a,\|\cdot\|_b$ are <Lipschitz equivalent norms> when constants $c,C>0$ satisfy $c\|x\|_a\leq\|x\|_b\leq C\|x\|_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a <Cauchy sequence> for one <norm> exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one <norm> is complete, a <Cauchy sequence> for the other converges using the first <norm> and hence also the second. Interchanging the <norms> proves \b[completeness is preserved in both directions].
Write $x=\sum_i x_ie_i$ and set $C=\sum_i\|e_i\|$. If $\|x\|_2=1$, then $|x_i|\leq1$, and the <triangle inequality> gives
$$
\|x\|\leq\sum_i|x_i|\|e_i\|\leq C.
$$
Scaling a nonzero $x$ gives $\|x\|\leq C\|x\|_2$ for every $x$. The reverse <triangle inequality> now implies
$$
\bigl|\|x\|-\|y\|\bigr|\leq\|x-y\|\leq C\|x-y\|_2,
$$
so $x\mapsto\|x\|$ is continuous, indeed Lipschitz, in the <Euclidean norm>.
Suppose there were no positive lower bound on the Euclidean <unit sphere>. We could choose $x_j$ with $\|x_j\|_2=1$ but $\|x_j\|<1/j$. The <Bolzano-Weierstrass theorem> supplies a subsequence converging in the <Euclidean norm> to $x$. Continuity of both <norms> gives $\|x\|_2=1$ and $\|x\|=0$, contradicting definiteness. Therefore some $m>0$ satisfies $\|x\|\geq m$ on the unit sphere. Scaling again proves the <finite-dimensional equivalence of norms>:
$$
\boxed{m\|x\|_2\leq\|x\|\leq\left(\sum_{i=1}^n\|e_i\|\right)\|x\|_2\quad\text{for every }x\in\mathbb R^n.}
$$
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