I was wondering if for $x_i\in\mathbb{R}_{\geq0}$ the inequality $$ \left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} x_{i} $$ Holds. If so, is there a name for it?
My attempt
$$ \sum_i^n i = \frac{n(n+1)}{2} $$ $$ \sum_i^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ Since the right hand side is $O(n^{3/2})$ and the left hand side is $O(n^2)$, and n is a positive integer, the inequality should hold.
Yes. It is the triangle inequality (for right triangles).
To explain. Consider a point $x=(x_1,x_2, \cdots, x_n) \in \mathbb{R}_{\ge 0}^n$.
$$|x|= \left(\sum_{i=1}^n x_i^2\right)^{1/2}$$ is the diagonal of a hyperparallelapiped. The sum of the "legs" is $\sum_{i=1}^n x_i$.
We can prove it by induction on $n$.
$$\sqrt{x_1^2 + x_2^2} \le x_1+x_2. \quad (*)$$ Also known as $(x_1+x_2)^2=x_1^2+x_2^2+2x_1x_2 \ge x_1^2+x_2^2$.
Assume $$\sqrt{x_1^2+ \cdots x_{n-1}^2} \le \sum_{i=1}^{n-1} x_i$$
Let $y=(x_1,x_2,\cdots,x_{n-1})\in \mathbb{R}_{\ge 0}^{n-1}$.
$$|y|= \left(\sum_{i=1}^{n-1} x_i^2\right)^{1/2}$$
Now apply $(*)$ to $\{|y|,x_n\}$.
I don't know whether there's a name for it, but note that\begin{align}\left(\sum_{i=1}^nx_i\right)^2&=\sum_{i=1}^nx_i^{\,2}+\overbrace{\sum_{i\ne j}x_ix_j}^{\geqslant0}\\&\geqslant\sum_{i=1}^nx_i^{\,2}\end{align}and that therefore$$\sum_{i=1}^nx_i\geqslant\sqrt{\sum_{i=1}^nx_i^{\,2}}.$$
