Prove that $|a+b|^r\leq 2^{r-1}(|a|^r+|b|^r)$

Question

In my textbook, the inequality $$|a+b|^r\leq 2^{r-1}(|a|^r+|b|^r)$$ for any real number $a$, $b$ and $r\geq 1$, is called "elementary". It is not elementary to me, how do you prove it? I see that the inequality reduces to the triangle inequality if $r=1$ so I am more interested in $r>1$.

What is $r$? A natural number? A rational number? A real number? — James Garrett, Oct 31 '17 at 20:28
For $r = 1$, the inequality reduces to what is called the triangle inequality: $| a + b | \leq |a| + |b|$. You could prove it by cases on whether $a$ and $b$ are positive or negative. For general $r$, you could proceed by induction. — Austin Mohr, Oct 31 '17 at 20:29
@AustinMohr Thanks, I've tried induction with no success (it grows into too many cases). — Elias, Oct 31 '17 at 20:38

score 2 · Answer 1 · answered Oct 31 '17 at 20:48

2

Also, by Holder $$\left(|a|^r+|b|^r\right)(1+1)^{r-1}\geq\left(\left(|a|^r\cdot1^{r-1}\right)^{\frac{1}{1+r-1}}+\left(|b|^r\cdot1^{r-1}\right)^{\frac{1}{1+r-1}}\right)^{1+r-1}=\left(|a|+|b|\right)^r$$

answered Oct 31 '17 at 20:48

Michael Rozenberg

203,855

score 1 · Accepted Answer · answered Oct 31 '17 at 20:37

1

Since $x\mapsto |x|^r$ is convex when $r\geq 1$, Jensen's inequality implies $$\left| \frac{a+b}2\right|^r \leq \frac 12 |a|^r + \frac 12 |b|^r$$

which easily transforms into what you're looking for .

answered Oct 31 '17 at 20:37

Gabriel Romon

36,881

Ah, so easy! Thanks! – Elias Oct 31 '17 at 20:37

Raffaele · Answer 3 · 2017-11-01T15:36:55.960

0

Proof by induction

for $r=1$ we have triangular inequality so inequality is true

suppose it holds for $r>1$ $$|a+b|^r\leq 2^{r-1}(|a|^r+|b|^r)\quad I.H.$$ rewrite it as $$|a+b|^r\leq \frac12\left((|2a|^r+|2b|^r)\right)\to 2|a+b|^r\leq |2a|^r+|2b|^r$$ we must show it is true for $r+1$ $$2|a+b|^{r+1}=2|a+b||a+b|^r\leq^{I.H.} |a+b|\left(|2a|^r+|2b|^r\right)\leq \left(|a|+|b|\right)\left(|2a|^r+|2b|^r\right)=$$ $$=|2a|^{r+1}+|a||2b|^r+|b||2a|^r+|2b|^{r+1}\geq |2a|^{r+1}+|2b|^{r+1}$$ so we have proved that $$2|a+b|^{r+1}\leq |2a|^{r+1}+|2b|^{r+1}$$

and the proof is complete.

Hope it helps

$$...$$

edited Nov 01 '17 at 15:36

answered Oct 31 '17 at 21:02

Raffaele

26,731

"If $r$ is not integer then as $\lfloor r\rfloor<r<\lceil r \rceil$ and $x^r$ is increasing for $r>1$ it can be shown that inequality holds for any $r$" Hmmm... How would you show it for $r=2.5$, say, using that it holds for $r=2$ and for $r=3$? – Did Oct 31 '17 at 22:11
@Did Thank you for your comment. I edited my answer. If you approve it can you please cancel your downvote? – Raffaele Nov 01 '17 at 11:30
Did I misread or did you deduce $$u\leqslant w$$ from $$u\leqslant v\geqslant w\ ?$$ (Additional suggestion: Try to concentrate much more on the mathematical accuracy of your answers (whose lack of is, frankly speaking, rather embarrassing at present), than on knowing who votes what for which reason, something that is none of your business. – Did Nov 01 '17 at 13:33
And a last (for me) remark: your answer received an upvote although even you might agree that it is seriously wrong. Why are you not bothered by this inappropriate and misleading upvote, but exclusively by the (appropriate) downvote your answer also received? – Did Nov 01 '17 at 13:36

Prove that $|a+b|^r\leq 2^{r-1}(|a|^r+|b|^r)$

3 Answers3