Inequality $\left|\,x_1\,\right|+\left|\,x_2\,\right|+\cdots+\left|\,x_p\,\right|\leq\sqrt{p}\sqrt{x^2_1+x^2_2+\cdots+x^2_p}$

This is an application of Jensen's Inequality: $$ \left(\frac1p\sum_{k=1}^p|x_k|\right)^2\le\frac1p\sum_{k=1}^p|x_k|^2 $$ since $f(x)=x^2$ is convex.

I imagine two possible ways to solve this: with the help of the QM-AM inequality, and with Cauchy-Schwarz inequality.

Using the QM-AM inequality, we see that: $$\frac{|x_1|+|x_2| +... + |x_p|}{p} \le \sqrt{\frac{|x_1|^2+|x_2|^2 +... + |x_p|^2}{p}}$$ Multiplying both sides by $p$: $$\begin{align}|x_1|+|x_2| +... + |x_p| &\le p\sqrt{\frac{|x_1|^2+|x_2|^2 +... + |x_p|^2}{p}}\\ &=\sqrt{p}\sqrt{|x_1|^2 + |x_2|^2 +... + |x_p|^2}\\ &=\sqrt{p}\sqrt{x_1^2 + x_2^2 +... + x_p^2}\end{align}$$

proving the desired statement.

Using the alternative Cauchy-Schwarz inequality (which really is, in some sense, a generalization of the QM-AM-GM-HM inequalities), we get :

$$(|x_1|\cdot1 + |x_2|\cdot1 +...+|x_p|\cdot1)^2 \le (|x_1|^2 + |x_2|^2 +... + |x_p|^2)(\underbrace{1 + 1 + ... + 1}_\text{$p$})$$ or: $$(|x_1|+|x_2| +... + |x_p|)^2 \le (|x_1|^2 + |x_2|^2 +... + |x_p|^2)(p)$$

Taking square roots of both sides, we get : $$|x_1|+|x_2| +... + |x_p| \le \sqrt{p}\sqrt{|x_1|^2 + |x_2|^2 +... + |x_p|^2}$$

But since we know that for all real $x$, $|x|^2 = x^2$, we can reduce this to:

$$|x_1|+|x_2| +... + |x_p| \le \sqrt{p}\sqrt{x_1^2 + x_2^2 +... + x_p^2}$$