I could use some help with proving this inequality:

$$\left|\,x_1\,\right|+\left|\,x_2\,\right|+\cdots+\left|\,x_p\,\right|\leq\sqrt{p}\sqrt{x^2_1+x^2_2+\cdots+x^2_p}$$ for all natural numbers p.

Aside from demonstrating the truth of the statement itself, apparently $\sqrt{p}$ is the smallest possible value by which the right hand side square root expression must be multiplied by in order for the statement to be true. I've tried various ways of doing this, and I've tried to steer clear of induction because I'm not sure that's what the exercise was designed for (from Bartle's Elements of Real Analysis), but the best I've been able to come up with is proving that the statement is true when the right hand side square root expression is multiplied by p, which seems pretty obvious anyway. I feel like I'm staring directly at the answer and still can't see it. Any help would be appreciated.

  • 6
    $\begingroup$ Have you learned the Cauchy-Schwarz inequality? $\endgroup$ Commented Feb 16, 2014 at 3:12

2 Answers 2


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:

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

  • $\begingroup$ This is good. Thank you. I can clearly see it now. $\endgroup$
    – mrmingus
    Commented Feb 16, 2014 at 4:16
  • $\begingroup$ @mrmingus I've edited my answer to include the QM-AM inequality, which is perhaps sufficient for proving the statement. No problem! If you are satisfied with my answer, please remember to accept my answer (the tick to the left of my answer). $\endgroup$
    – Yiyuan Lee
    Commented Feb 16, 2014 at 4:20
  • $\begingroup$ I have not heard of the QM-AM inequality. I will check this method out as an alternative solution. Thanks again. $\endgroup$
    – mrmingus
    Commented Feb 16, 2014 at 4:22

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