Proving $|x|$ is a norm in $\mathbb {R}^n$

I need some direction on how to start on showing that $| x+y|\leq|x|+|y|$ in $\mathbb R^n$. Note that $$|x|=\left(\sum\limits_{j=0}^n x_i^2\right)^{1/2}$$ Thank you, Klara

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You say "absolute value" in the title of your question. Are you referring to the Euclidean norm, or the taxicab norm? – wj32 Nov 3 '12 at 3:27
@wj32 Euclidean norm – Klara Nov 3 '12 at 3:32

It is suffice to use the cauchy schwarz inequality. more precisely

$$\|x+y\|^2=\langle x+y, x+y\rangle=\langle x, x\rangle+2\langle x, y\rangle+\langle y, y\rangle,$$ now using the cauchy-schwarz inequality, that is $|\langle x, y\rangle|\leq \langle x,x\rangle^{1/2}\langle y, y\rangle^{1/2}$ we have

$$\|x+y\|^2\leq \langle x, x\rangle +2\langle x,x\rangle^{1/2}\langle y, y\rangle^{1/2}+\langle y,y\rangle= (\|x\|+\|y\|)^2,$$ now take the square root to both side and you got it.

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Start with $$\Vert x+y \Vert^2=(x+y)\cdot(x+y)=\Vert x \Vert^2 + 2(x \cdot y) + \Vert y \Vert^2$$ where $\cdot$ is the dot product, and apply an inequality you know relating $x \cdot y$, $\Vert x \Vert$ and $\Vert y \Vert$.

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Yes, that's what I'm using: $\Vert x \Vert = \sqrt{\sum_i x_i^2}$ – wj32 Nov 3 '12 at 3:37
Are you objecting to the use of the double vertical bars ($\Vert \cdot \Vert$)? Feel free to replace $\Vert$ with $|$ if that's the notation you're more comfortable with. – wj32 Nov 3 '12 at 3:40
I don't know what you're objecting to. In your first comment you say that your definition is $|x|=\sqrt{\sum_{i=0}^n x_i^2}$. I then confirmed that my answer uses the same definition. – wj32 Nov 3 '12 at 3:44
Instead of throwing a bunch of equations at me, it would be more constructive if you carefully state what the problem is. An example would be "I define the Euclidean norm as ... but your answer uses ...". – wj32 Nov 3 '12 at 3:52
I don't appreciate you getting upset. – Klara Nov 3 '12 at 4:17

It seems from some comments you made on the other answers that you're getting confused by the notation. Let me try to show you the idea in an easy case. Lets concentrate on the case in which $n = 1$, so we're working with $\mathbb{R}$.

So for real numbers $x, y \in \mathbb{R}$, you want to show that $|x + y| \leq |x| + |y|$ where $|x|$ is just the usual absolute value now since $\sqrt{x^2} = |x|$ in this case.

Now, observe that it suffices to prove that $|x + y|^2 \leq (|x| + |y|)^2$. So lets try to work backwards from here. We have the following.

$$\begin{eqnarray} |x + y|^2 \leq \color{red}{(|x| + |y|)^2} \iff (x + y)^2 \leq \color{red}{|x|^2 + 2 |x||y| + |y|^2} \\ \iff x^2 + 2xy + y^2 \leq |x|^2 + 2 |x||y| + |y|^2\\ \iff 2xy \leq 2|x||y|\\ \iff xy \leq |x||y| \end{eqnarray}$$

Now, this last inequality is the key step in the proof, because all the other steps were just squaring things out and using the fact that $x^2 = |x|^2$ to cancel some terms out.

In the general case, for any $n$, the idea of the proof is exactly the same, as you can see from the other answers. But now the key step, or the key inequality, becomes what is called the Cauchy-Schwarz inequality as already indicated in another answer.

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