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Is there $C=C(p)$ constant that depend only on $p$ such that if $a,b > 0$ we have $$ (a +b)^{p} \le C(a^{p} + b^{p})? $$ where $p \in \mathbb{N}$ is fixed. For example, if $p=2$ $$ (a+b)^{2} \le 2(a^{2} + b^{2}). $$

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    Yes, on $\mathbb R^2$ are all p-adic norms equivalent.2012-05-26

2 Answers 2

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$C(p)=2^{p-1}$ works for any $p\in [1,\infty[.$

Infact by the convexity of $t\in[0,\infty[\to t^p\in[0,\infty[,$ for any $a,b\in[0,\infty[,$ you get $$(a+b)^p=2^p\left(\frac{1}{2}a+\frac{1}{2}b\right)^p\leq 2^p\left(\frac{1}{2}a^p+\frac{1}{2}b^p\right),$$

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    And it's the best possible $a=b=1$.2012-05-26
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Without loss of generality we may assume $0

  • If $p\ge1$, then $g'(x)\le0$ on $[0,\infty)$, $g$ is decreasing on $[0,\infty)$ and $C_p=g(1)=2^{p-1}$, as shown in Giuseppe's answer.
  • If $0
  • If $p=0$ then we have equality with $C_0=1/2$.
  • If $p<0$, then $g'(x)\le0$ on $[0,\infty)$, $g$ is decreasing on $[0,\infty)$ and $C_p=g(1)=2^{p-1}$.