# Proving norm inequality

There is a brief proof in my textbook that I have one question about.

We are supposed to prove that $||x||_{1} \leq n||x||_{\infty}$ for $x \in \mathbb{R}^n$

The book writes the following:

$||x||_{1} = \sum_{i=1}^{n} |x_i| \leq \sum_{i=1}^{n}\{\max_{1 \leq j \leq n} |x_j| \} \leq \sum_{i=1}^{n} ||x||_{\infty} = n||x||_{\infty}$

The one thing I don't quite follow is when the book writes:

$\sum_{i=1}^{n}\{\max_{1 \leq j \leq n} |x_j| \} \leq \sum_{i=1}^{n} ||x||_{\infty}$. In my book the definition of $||x||_{\infty}$ is given as:

$$||x||_{\infty} = \max_{1 \leq j \leq n}|x_j|$$

So shouldn't the inequality $\sum_{i=1}^{n}\{\max_{1 \leq j \leq n} |x_j| \} \leq \sum_{i=1}^{n} ||x||_{\infty}$ actually be an equality?

I would really appreciate it if someone could explain this to me!

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Short answer: Yes. – cardinal Sep 8 '12 at 19:25
Longer answer: it would also be correct with equality, but it’s not incorrect as it stands. After all, if $a=b$, then it’s also true that $a\le b$. – Brian M. Scott Sep 8 '12 at 19:26
Thank you for the answers! I sometimes get a bit confused about proofs when they switch between inequalities and equalities, so I appreciate the clarification! – Kristian Sep 8 '12 at 19:28
@Brian: Would you like to make that an answer so that it can be accepted? – Ilmari Karonen Sep 8 '12 at 19:58
@Ilmari: Done. $\;$ – Brian M. Scott Sep 8 '12 at 20:01

It would also be correct with equality, but it’s not incorrect as it stands. After all, if $a=b$, then it’s also true that $a\le b$, and the inequality is all that’s actually needed here.