(adsbygoogle = window.adsbygoogle || []).push({}); "1-norm" is larger than the Euclidean norm

Define, for each [itex]\vec{x} = (x_1, \ldots, x_n) \in \mathbb{R}^n[/itex], the (usual) Euclidean norm [itex]\Vert{\vec{x}}\Vert = \sqrt{\sum_{j = 1}^n x_j^2}[/itex] and the 1-norm [itex]\Vert{\vec{x}}\Vert_1 = {\sum_{j = 1}^n |x_j|}[/itex].

How can we show that, for all [itex]\vec{x} \in \mathbb{R}^n[/itex], we have [itex]\Vert{\vec{x}}\Vert \leq \Vert{\vec{x}}\Vert_1[/itex]?

I'm thinking of writing [itex]\Vert{\vec{x}}\Vert^2 \leq \Vert{\vec{x}}\Vert_1^2[/itex] and then showing (probably inductively) that the sum of squares of (finitely) many numbers is not larger than the square of the sum of the absolute values of the same numbers; i.e. show [itex]{\sum_{j = 1}^n x_j^2} \leq (\sum_{j = 1}^n |x_j|)^2[/itex] by induction on n. For n = 1 and 2 this is simple enough. The inductive step is tricky, but I feel like using an induction argument is totally overdoing it.

I ask this because I read that this is trivial, but I don't see it immediately. Do you?

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# 1-norm is larger than the Euclidean norm

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