# $\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} x_{i}$

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.

• Why on earth should the equality hold in general just because it holds for one extremely special case (the case you mentioned)? – user21820 Jan 13 at 8:26
• @user21820, looks like a typo (that was edited). The word "inequality" was cut (now pasted back together). – mjw Jan 13 at 14:44
• @mjw: I meant "inequality" in my comment, and my objection is correct when that typo is fixed. – user21820 Jan 13 at 15:22
• You are writing about the "attempt" not the actual problem statement, right? The inequality is true in the statement. – mjw Jan 13 at 18:08

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$$\sum_{i=1}^nx_i\geqslant\sqrt{\sum_{i=1}^nx_i^{\,2}}.$$

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 $$\{|y|,x_n\}$$.

• Looks like somebody didn't understand this answer. Sorry about that. – mjw Jan 12 at 23:25
• Was not me! +1. – Blade Jan 12 at 23:37
• @Blade, didn't mean you. Anyway, will try to write more clearly going forward $\cdots.$ – mjw Jan 12 at 23:39
• Neutralizing +1 for a geometric approach. – J.G. Jan 13 at 22:24