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

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

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.