I try to prove $\|a+b\|_2\le\|a\|_2+\|b\|_2$.Where$\|\cdot\|_2=(\sum |x_i|^2)^{1/2}$.a,b can be bounded sequences or n-dimentional vectors take value on $\mathbb{C}$.
My first try is to see if for each term, $|a_i+b_i|^2\le|a_i|^2+|b_i|^2$. This seems to be wrong when $ab<0$.
I know that $|a+b|\le|a|+|b|$ and $|\sum _{1}^{n}a_ib_i|\le(\sum _{1}^{n}|a_i|^2)^{1/2}(\sum _{1}^{n}|b_i|^2)^{1/2} $. But I am not sure how to prove when the right hand side is addition instead of product.
And I think I don't quite understand the general logic behind these inequalities, so I always have troubles proving similar inequalities. Could anyone tell me some good english reference books about this topic?
$$\Vert a+b \Vert^2 = \langle a+b,a+b \rangle = \langle a,a \rangle +\langle b,b \rangle +\langle a,b \rangle + \langle b,a \rangle $$ By Cauchy-Schwartz inequality, $$ \leq \Vert a \Vert^2+\Vert b \Vert^2 +\Vert a \Vert \Vert b \Vert+\Vert a\Vert \Vert b \Vert \leq \left(\Vert a \Vert+\Vert b \Vert \right)^2 $$
Hence $\Vert a+b \Vert \leq \Vert a \Vert + \Vert b \Vert$
This bound will be met when $\langle a,b \rangle = \Vert a \Vert \Vert b \Vert$. This happens when $a = \lambda b$ with $\lambda \geq 0$ according to a corollary to Cauchy-Schwartz inequality.