Generalization of $(a+b)^2\leq 2(a^2+b^2)$

Question

We know that, $(a+b)^2\leq 2(a^2+b^2)$. Do we have anything similar for $$\left(\sum_{i=1}^N a_i\right)^2.$$ where $a_i\in \mathbb{R}\ \ \ \ \forall\ i\in \{1,\cdots,N\}$.

For $n=3$, we get \begin{equation} \begin{aligned} (a_1+a_2+a_3)^2&\leq 2\left((a_1+a_2)^2+a_3^2\right) \\&\leq 2\left(2(a_1^2+a_2^2)+a_3^2 \right). \end{aligned} \end{equation} Do we have some sort of generalization?

Answer 1

It's C-S: $$n(a_1^2+a_2^2+...+a_n^2)=$$ $$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)\geq(a_1+a_2+...+a_n)^2.$$

Answer 2

The generalization is

$$\left(\sum_{i=1}^N a_i\right)^2\le N\sum_{i=1}^N a_i^2$$ which degenerates to equality if all $a_i$ are equal. It is the Cauchy-Schwarz inequality applied to vectors $(1,1,\dots,1)$ and $(a_1,a_2,\dots,a_n)$.

Answer 3

Others have mentioned this formula already but I want to remark that we can strengthen it a bit to $$ |a_1|+\cdots+|a_n| \le \sqrt{n}\,\left(a_1^2+\cdots+a_n^2 \right)^{1/2}. $$ Moreover, the "other direction" we also have $$ \left(a_1^2+\cdots+a_n^2 \right)^{1/2} \le |a_1|+\cdots+|a_n|, $$ which follows from super-additivity of $x\mapsto x^2$ on $[0,\infty)$. Together they show that these two norms on $\Bbb R^n$ ($||\cdot||_1$ and $||\cdot||_2$) are equivalent.