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 $r>1$.

Answer 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 .

Answer 2

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

Answer 3

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

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