Proving that $l^2$ is a vector space - how is the following true? In part of my lecture notes, the lecturer states that: $$l^2 := \big\{ x = (x_i)_{i \in \mathbb{N}} \ \ \big| \ \ \sum_{i=1}^{\infty} |x_i|^2 < +\infty \big\}$$ is a vector space since: $$|x_i + y_i|^2 \leq 2|x_i|^2 + 2|y_i|^2$$ I understand why, if that is true, then $l^2$ is a vector space (since scalar multiplication is not hard to justify and if you take the sum over $i$ of the above inequality then you can see that the right hand side consists of terms which are already contained in $l^2$).
However I do not understand why that inequality is true. It might be easy to prove that it's true in the case of $2$, but what about for general $p$? For example, can I then say: $$|x_i + y_i|^p \leq p|x_i|^p + p|y_i|^p$$ in order to prove that $l^p$ is a vector space? If so then how do I know that's true (it might not be, I'm just eager to know)?
Thanks for any advice you can give me!
 A: The case $p=2$:
$|x_i+y_i|^2=|(x_i+y_i)^2|=|x_i^2+2x_iy_i+y_i^2| \le |x_i|^2+2|x_i||y_i|+|y_i|^2 \le 2|x_i|^2+2|y_i|^2$.
The last inequality is valid since $2|x_iy_i| \le |x_i|^2+|y_i|^2$ (Binomi !)
The general case: in this case the correct inequality is
$|x_i+y_i|^p \le 2^{p-1}(|x_i|^p+|y_i|^p)$
A: The inequality $\;\lvert x_i+y_i\rvert^p \le 2^{p-1}(\lvert x_i\rvert^p+\lvert y_i\rvert^p)$ comes from Jensen's inequality for the convex function $t\mapsto t^p\;$ ($p>1)$:
$$\biggl(\frac{\lvert x+y\rvert}2\biggr)^p\le\biggl(\frac{\lvert x\rvert+\lvert y\rvert}2\biggr)^p \le\frac{\lvert x\rvert^p+\lvert y\rvert^p}2.$$
A: in general $l^2 = \{ (x_n)_n, x_n \in \mathbb{R}\}$ is proven to be a Hilbert space  by defining the inner product
$$\langle x,y \rangle = \sum_n x_n y_n$$ which is bilinear and hence $$\|x+y\|^2 = \langle x+y,x+y \rangle=\langle x,x \rangle+\langle y,y \rangle+\langle y,x \rangle+\langle x,y \rangle = \|x\|^2+\|y\|^2+ 2\langle x,y \rangle$$
Now $|\langle x,y \rangle| \le \|x\| \|y\|$ is the Cauchy-Schwartz inequality, that is obvious when writing that $y = ax+z, a =\frac{\langle x,y\rangle}{\|x\|^2}, z = y-ax, \langle x,z\rangle = 0$  so that $\|y\|^2 = |a|^2\|x\|^2+ \|z\|^2$ and $|\langle x,y\rangle| = |a| \|x\|^2 \le \|x\|\, \|y\|$
and overall
$$\|x+y\|^2 = \|x\|^2+\|y\|^2+2 \langle x,y\rangle < \|x\|^2+\|y\|^2+2 \|x\| \, \|y\| = (\|x\|+\|y\|)^2$$
