TITLE: Generalization of $(a+b)^2\leq 2(a^2+b^2)$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.