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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. –  Simon Markett May 26 '12 at 18:36

2 Answers 2

up vote 2 down vote accepted

$C(p)=2^{p-1}$ 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 $a=b=1$. –  Davide Giraudo May 26 '12 at 19:00

Without loss of generality we may assume $0<a\le b$. Let $x=b/a$. Then the sought for inequality is equivalent to $$ (1+x)^p\le C_p(1+x^p)\qquad\forall x\ge1. $$ We can take $$C_p=\sup_{x\ge1}\frac{(1+x)^p}{1+x^p}.$$ Let $g(x)=\dfrac{(1+x)^p}{1+x^p}$. Then $$ g'(x)=p\,\frac{(1+x)^{p-1}}{(1+x^p)^2}\bigl(1-x^{p-1}\bigr). $$

  • 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 $0<p<1$, then $g'(x)\ge0$ on $[0,\infty)$, $g$ is increasing on $[0,\infty)$ and $C_p=\lim_{x\to\infty}g(x)=1$.
  • 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 $C_p=g(1)=2^{p-1}$.
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