My goal is to have a chain of inequalities so that $|\sum_{i=1}^{n}a_ix_i|\le \ldots \le c\sqrt{\sum_{i=1}^{n}x_i^2}$
$|\sum_{i=1}^{n}a_ix_i|\le\sum_{i=1}^{n}|a_ix_i|=\sum_{i=1}^{n}\sqrt{(a_ix_i)^2}\le\sqrt{\sum_{i=1}^{n}(a_ix_i)^2}\le\sqrt{\max_{i=1,\ldots,n}a_i^2\sum_{i=1}^{n}x_i^2}=\max_{i=1,\ldots,n}|a_i|\sqrt{\sum_{i=1}^{n}x_i^2}$
Is this correct?
Jensen inequality means for a convex function, $f$, we have
$$f(E(X)) \le E(f(X))$$
square root is a concave function.
$$\frac{\sum_{i=1}^n\sqrt{(a_ix_i)^2}}{n} \le \sqrt{\frac{\sum_{i=1}^n(a_ix_i)^2}{n}}$$
Hence we have
$$\sum_{i=1}^n\sqrt{(a_ix_i)^2}\le \sqrt{n}\sqrt{\sum_{i=1}^n(a_ix_i)^2}$$
To see that $\sqrt{n}$ can't be omitted, let $a_i=1, n=2, x_1=3, x_2=4$.
Without $\sqrt{n}$, we have $$3+4 \le \sqrt{3^2+4^2}$$ which is not true.
Edit:
By Cauchy-Schwarz,
$$|\langle a, x \rangle | \le \|a\|_2 \|x\|_2$$