Is $|x_1|+|x_2|+...+|x_n| \leq \sqrt{x_1^2+x_2^2+...+x_n^2}$?

Question

$$|x_1|+|x_2|+...+|x_n| \leq \sqrt{x_1^2+x_2^2+...+x_n^2}$$

This inequality is wrong has

$$\sqrt{x_1^2+x_2^2+...+x_n^2}\leq \sqrt{(x_1+x_2+...+x_n)^2}\leq|x_1+x_2+...+x_n|\leq |x_1|+|x_2|+...+|x_n|$$

But where in the process I got it wrong?

$$|x_1|+|x_2|+...+|x_n|=\sqrt{x_1^2}+\sqrt{x_2^2}+...+\sqrt{x_n^2}\leq \sqrt{x_1^2+x_2^2+...+x_n^2}$$

But why would the original inequality be true? This is not Cauchy—Schwarz. — Clement C., Nov 08 '17 at 22:21
@ClementC. you mean why $\sqrt{x_1^2+x_2^2+...+x_n^2}\leq|x_1|+|x_2|+...+|x_n|$ is true? — gbox, Nov 08 '17 at 22:23
@gbox No, I don't mean that. I mean "why would the OP originally believe the inequality $\sum_i \lvert x_i\rvert \leq \sqrt{\sum_i x_i^2}$ to hold in the first place?" This question does not come out of nowhere, so what prompted it? — Clement C., Nov 08 '17 at 22:24
@ClementC. Just came across my mind when tried to prove the original and correct inequality and wanted to know why it is wrong — gbox, Nov 08 '17 at 22:26
@gbox the correct inequality you point is basically the monotonicity of $\ell_p$ norms: $\lVert x\rVert_p \leq \lVert x\rVert_q$ whenever $0< q\leq p \leq \infty$. — Clement C., Nov 08 '17 at 22:26

score 6 · Accepted Answer · answered Nov 08 '17 at 22:20

6

Try $x_i=1$.

You'll get $$n\leq\sqrt{n},$$ which is wrong for all $n>1$.

answered Nov 08 '17 at 22:20

Michael Rozenberg

score 2 · Answer 2 · answered Nov 08 '17 at 22:20

2

This part is not correct:

$$ \sqrt{x_1^2+x_2^2+...+x_n^2}\leq \sqrt{(x_1+x_2+...+x_n)^2}$$

Let $x_1 = 3, x_2 = -4$. Then $\sqrt{3^2 + (-4)^2} = 5 > 1 = \sqrt{(3 + (-4))^2}$

answered Nov 08 '17 at 22:20

Larry B.

Replace all $x_i$'s by $\lvert x_i\rvert$, the OP's inequalities become correct and contradict the purported (and wrong) original inequality. – Clement C. Nov 08 '17 at 22:22

score 0 · Answer 3 · answered Feb 07 '24 at 09:13

To complete the previous answer given we can note that: $ |x_1+x_2+...+x_n| = |x_1|+|x_2|+...+|x_n| \Leftrightarrow $ all the $x_i$ have the same sign.
Prove $(\Rightarrow)$:
1-We define $A= \sum_{\left \{ x_i : x_i \geq 0 \right \}}x_i $ (that means the sum of all the different positive or equal to zero $x_i$ ) and $B=\sum_{\left \{ x_i : x_i < 0 \right \}}x_i$ (that means the sum of all the different striclty negative $x_i$ ).
2-Now we can writte by hypothesis $|A+B| = |A|+|B| = A +|B| $. On an other side we can note that $A+B=A-|B|$ (because $B$ is necessarily negative) hence we have.
3-So we have $A +|B|=|A+B|=|A-|B|| \Rightarrow (A +|B|)^2-(|A+B|)^2=0$ and $(A +|B|)^2-(|A+B|)^2=4AB$, thus $4AB=0$ and so or $A=0, B \neq 0$ in such a case it means that the sum $x_1+x_2+...+x_n$ is a sum of only negative numbers. Or $A \neq 0, B = 0$ in such a case it means that the sum $x_1+x_2+...+x_n$ is a sum of only positive numbers. And the case $A=0=B$ is a trivial case.
Prove $(\Leftarrow)$: trivial

Q.E.D.

3 Answers3