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 $p=2$ $$ (a+b)^{2} \le 2(a^{2} + b^{2}). $$
natural question of inequality
2
$\begingroup$
inequality
-
1Yes, on $\mathbb R^2$ are all p-adic norms equivalent. – 2012-05-26
2 Answers
2
$C(p)=2^{p-1}$ 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 $$(a+b)^p=2^p\left(\frac{1}{2}a+\frac{1}{2}b\right)^p\leq 2^p\left(\frac{1}{2}a^p+\frac{1}{2}b^p\right),$$
-
1And it's the best possible $a=b=1$. – 2012-05-26
0
Without loss of generality we may assume $0