[proofplan] We compare the two quantities by squaring, since both norms are nonnegative. The lower bound follows because each coordinate square is one summand in the Euclidean sum, so each coordinate absolute value is bounded by $|x|$, and then we take the maximum over the finitely many coordinates. The upper bound follows because every coordinate absolute value is bounded by the maximum coordinate absolute value, so the sum of the coordinate squares is bounded by $n|x|_\infty^2$. [/proofplan] [step:Bound each coordinate by the Euclidean norm] Fix $x=(x_1,\dots,x_n)\in\mathbb{R}^n$. For each index $i\in\{1,\dots,n\}$, define the nonnegative real number \begin{align*} S := \sum_{j=1}^n x_j^2. \end{align*} Since every summand $x_j^2$ is nonnegative, we have \begin{align*} x_i^2 \leq \sum_{j=1}^n x_j^2 = S. \end{align*} The square-root function is increasing on $[0,\infty)$, and both $x_i^2$ and $S$ belong to $[0,\infty)$. Therefore \begin{align*} |x_i| = (x_i^2)^{1/2} \leq S^{1/2} = |x|. \end{align*} Since this holds for every $i\in\{1,\dots,n\}$, taking the maximum over $i$ gives \begin{align*} |x|_\infty = \max_{1\leq i\leq n}|x_i| \leq |x|. \end{align*} [guided] Fix $x=(x_1,\dots,x_n)\in\mathbb{R}^n$. The goal of this step is to prove that the largest coordinate absolute value cannot exceed the Euclidean norm. Define \begin{align*} S := \sum_{j=1}^n x_j^2. \end{align*} This number is nonnegative because every square $x_j^2$ is nonnegative. For a fixed index $i\in\{1,\dots,n\}$, the term $x_i^2$ appears as one of the summands in $S$, so \begin{align*} x_i^2 \leq \sum_{j=1}^n x_j^2 = S. \end{align*} Now both sides are nonnegative, so we may take square roots while preserving the inequality. This gives \begin{align*} |x_i| = (x_i^2)^{1/2} \leq S^{1/2}. \end{align*} By the definition of the Euclidean norm in the statement, $S^{1/2}=|x|$. Hence \begin{align*} |x_i| \leq |x| \end{align*} for every $i\in\{1,\dots,n\}$. Since $|x|_\infty$ is the maximum of the finite set $\{|x_1|,\dots,|x_n|\}$, and every member of this set is at most $|x|$, the maximum is also at most $|x|$: \begin{align*} |x|_\infty = \max_{1\leq i\leq n}|x_i| \leq |x|. \end{align*} [/guided] [/step] [step:Bound the Euclidean norm by the supremum norm] For each $i\in\{1,\dots,n\}$, the definition of the maximum gives \begin{align*} |x_i| \leq |x|_\infty. \end{align*} Squaring both sides, which are nonnegative, yields \begin{align*} x_i^2 = |x_i|^2 \leq |x|_\infty^2. \end{align*} Summing this inequality over $i\in\{1,\dots,n\}$ gives \begin{align*} \sum_{i=1}^n x_i^2 \leq \sum_{i=1}^n |x|_\infty^2 = n|x|_\infty^2. \end{align*} By the definition of $|x|$, this is \begin{align*} |x|^2 \leq n|x|_\infty^2. \end{align*} Both sides are nonnegative, so taking square roots gives \begin{align*} |x| \leq \sqrt{n}\,|x|_\infty. \end{align*} [/step] [step:Combine the two inequalities] The first step proved \begin{align*} |x|_\infty \leq |x|. \end{align*} The second step proved \begin{align*} |x| \leq \sqrt{n}\,|x|_\infty. \end{align*} Combining these inequalities gives \begin{align*} |x|_\infty \leq |x| \leq \sqrt{n}\,|x|_\infty. \end{align*} Since $x\in\mathbb{R}^n$ was arbitrary, the inequality holds for every $x\in\mathbb{R}^n$. [/step]