I want to show that the euclidean norm $$||\!\cdot\!||_2: \mathbb{R}^2\to \mathbb{R},x\mapsto \sqrt{x_1^2+x_2^2}$$
indeed is a norm. But I got stuck at the triangle inequality for which I cannot use the Cauchy Schwarz inequality. Instead I was given the two hints
i: $$\forall a,b\in \mathbb{R}: 2ab\leq a^2 + b^2 \Longleftrightarrow 0 \leq (a-b)^2$$
ii: $$\forall a\in \mathbb{R}, b\in \mathbb{R}_0^+ : a^2 \leq b^2 \Rightarrow a\leq b$$
So starting with $||x+y|| \leq ||x|| + ||y ||$ I have no clue where to use these hints.
I already tried substituting $a:=||x||,b:=||y||$ or squaring the inequality but I always end up with $2\cdot ||x+y|| \leq 2\cdot ||x||\cdot ||y||$.
That would be $2\cdot ||x+y|| \leq 2ab$ but it doesn't look right.
What clue am I missing?
Best Answer
I'd say squaring the inequality is a good idea. I'll just use $\Vert\cdot\Vert$ instead of $\Vert\cdot\Vert_2$. We have $$\Vert x+y\Vert^2=(x_1+y_1)^2+(x_2+y_2)^2=(x_1^2+x_2^2)+(y_1^2+y_2^2)+2(x_1y_1+x_2y_2)$$ and on the other hand $$(\Vert x\Vert+\Vert y\Vert)^2=\left(\sqrt{x_1^2+x_2^2}+\sqrt{y_1^2+y_2^2}\right)=(x_1^2+x_2^2)+(y_1^2+y_2^2)+2\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}.$$ So it really comes down to showing that $$x_1y_1+x_2y_2\leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$ (which is really just the Cauchy-Schwartz inequality). To show this inequality, try starting with $$0\leq (x_1y_2-x_2y_1)^2$$ and expanding.