Prove that this function define a norm on $\mathbb{R}^2$. Let $\| \;\|:\mathbb{R}^2\rightarrow\mathbb{R}$ be a function such that 
$ \| (x,y)\|=\sqrt{|x|^2+|y|^2}$ for all $(x,y)\in\mathbb{R}^2$.
I need to show that $\|\;\|$ defines a norm. I would like a hint in order to prove that $\parallel \; \parallel$ satisfies the triangle inequality:
$ \sqrt{|x_1+x_2|^2+|y_1+y_2|^2}\leq \sqrt{|x_1|^2+|y_1|^2}+\sqrt{|x_2|^2+|y_2|^2}$
Thanks!
 A: Let $x, y\in \mathbb{R}^2$. Then
\begin{align}
\|x+y\|^2&=\langle x+y, x+y\rangle\\&=\langle x, x\rangle + 2\langle x, y\rangle+\langle y, y\rangle \\ &\le \{\text{Cauchy-Schwarz}\} \\&\le \|x\|^2+2\|x\|\|y\|+\|y\|^2  \\
&=(\|x\|+\|y\|)^2,
\end{align}
which is your result.
A: You can do it directly, for $\mathbb{R}^2$; it becomes really complicated in higher dimension, where the Cauchy-Schwarz inequality is the best approach.
Square your inequality: you can because both terms are non negative. You can also remove the absolute values, because $(-a)^2=a^2$. Thus your inequality is equivalent to
$$
(x_1+x_2)^2+(y_1+y_2)^2\leq 
x_1^2+y_1^2+x_2^2+y_2^2
+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}.
$$
Expand the terms and you get, after rearranging terms,
$$
2(x_1x_2+y_1y_2)\le+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}
$$
that's obviously satisfied when $x_1x_2+y_1y_2<0$. So we proceed under the additional assumption that $x_1x_2+y_1y_2\ge0$, so the inequality is equivalent to the one we obtain by squaring (first remove the common factor $2$):
$$
(x_1x_2+y_1y_2)^2\le(x_1^2+y_1^2)(x_2^2+y_2^2)
$$
Expand the terms
$$
x_1^2x_2^2+2x_1x_2y_1y_2+y_1^2y_2^2\le
x_1^2x_2^2+x_1^2y_2^2+x_2^2y_1^2+y_1^2y_2^2
$$
and transport all terms to the right hand side
$$
0\le x_1^2y_2^2-2x_1x_2y_1y_2+x_2^2y_1^2
$$
which can be written
$$
0\le (x_1y_2-x_2y_1)^2
$$
that's certainly true.
