<s>[INST] Show that $||X||$ is a norm in $\mathbb{R}^2$ Let $a,b \in \mathbb{R}$ fixed numbers and the following norm in $\mathbb{R}^2$:
$$ ||x|| = \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}} \qquad x=(x_1,x_2)\in\mathbb{R}^2$$
Prove that it is, in fact, a norm.
Taking the rules of a norm, I get the following:


*

*$||x||\geq0$: since we take $x_1^2$ and $x_2^2$, it will never be negative.

*$||x||=0 \iff x=0$: same applies here. It is only $0$ if $x_1$ and $x_2$ are $0$.

*$||\lambda x||=|\lambda|||x||$: $||\lambda x||=\lambda \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}=\sqrt{ \frac{\lambda^2x_1^2}{a^2}+ \frac{\lambda^2x_2^2}{b^2}}=\sqrt{\lambda^2 (\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2})}=|\lambda|\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}=|\lambda| ||x||$

*$||x+y||\leq||x||+||y||:$The triangle inequality is the point where I got stuck. Not sure if $||x+y||= \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$ or $\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}+\sqrt{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$.
Any help or suggestion is appreciated!
 [/INST] Notice that $x+y=(x_1+y_1,x_2+y_2)$ so we have that $$||x+y||=\sqrt{\frac{(x_1+y_1)^2}{a^2}+\frac{(x_2+y_2)^2}{b^2}}$$ is the correct first step. You then need to manipulate it to show that it is greater than $$||x||+||y||=\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}+\sqrt{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$$
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