I've made proof by induction over $n$ for triangle inequality : $\left \| x+y \right \|_{e}\leq \left \| x \right \|_{e}+\left \| y \right \|_{e}$ ,where $\left \| x \right \|_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ for $x\in \mathbb{R}^{n}$. Is that proof also valid for triangle inequality for $\left \| x \right \|=\sqrt{\sum_{i=1}^{\infty}x_{i}^{2}}$ where $x\in \ell^2, \ell^2=\left\{{x\in\mathbb{R}^{\infty}}:\left \|x\right\|^{2}<\infty\right\}$ ? Maybe I should write down that proof, but I don't believe that induction proof could be also valid in case of infinite spaces.
$\left \| x+y \right \|_{e}\leq \left \| x \right \|_{e}+\left \| y \right \|_{e}$
$\sum_{i=1}^{n}(x_{i}+y_{i})^{2}\leq \sum_{i=1}^{n}(x_{i})^{2}+\sum_{i=1}^{n}(y_{i})^{2}+2\sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}$
$\sum_{i=1}^{n}x_{i}y_{i}\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}$
this is true when the true is that :
$\sum_{i=1}^{n}\left |x_{i}y_{i}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}$
above inequality is true for $n=1$ and we assume that it's true for $n$.
For $n+1$ we get : $\sum_{i=1}^{n}\left |x_{i}y_{i}\right |+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}+x_{n+1}^{2}\sum_{i=1}^{n}(y_{i})^{2}+y_{n+1}^{2}\sum_{i=1}^{n}(x_{i})^{2}+x_{n+1}^{2}y_{n+1}^{2}}$
using induction assumption we get :
$\sum_{i=1}^{n}\left |x_{i}y_{i}\right |+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}+x_{n+1}^{2}\sum_{i=1}^{n}(y_{i})^{2}+y_{n+1}^{2}\sum_{i=1}^{n}(x_{i})^{2}+x_{n+1}^{2}y_{n+1}^{2}}$
now we take cube of both sides and annihilate what we can, in result we get that 0=<(...+...)^2 so inequality is hold.
So is this proof valide for infite spaces ?