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Clearly if $a,b >0$ and $p \in \mathbb{N}$ $$ a^{p} + b^{p} \le (a+b)^{p} $$ Is there a constante $C = C(p)$ such that if $a,b >0$ and $p \in \mathbb{N}$ then \begin{equation} a^{p} - b^{p} \le C(p)(a-b)^{p} ? \end{equation}

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Does $C(p) = 1$ work? –  Thomas Jun 1 '12 at 16:38
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$a^{p} - b^{p} \le a^{p} + b^{p} \le (a+b)^{p}$ –  lhf Jun 1 '12 at 16:40
    
Sorry I would like $a^{p} - b^{p} \le C(p)(a-b)^{p} ?$ –  user29999 Jun 1 '12 at 16:45
    
@Marcos What lhf said is exactly that with $C(p)=1$. –  Alex Becker Jun 1 '12 at 16:48

2 Answers 2

up vote 8 down vote accepted

It is not possible to find a constant $C(p)$ such that

$$ a^{p} - b^{p} \le C(p)(a-b)^{p} $$

for all $a,b > 0$ when $p > 1$. For example, let $a = n+1$ and $b = n$. Then

$$ a^p-b^p = (n+1)^p - n^p = \sum_{k=0}^{p-1} \binom{p}{k} n^k \to \infty $$

but $a-b = 1$.

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Hint: Suppose that $p>1$. Let $b$ be huge and let $a=b+1$.

Added: One might argue that choosing $a-b=1$ is unfair. However, since the proposed inequality is homogeneous, it really makes no difference. For let $b=kx$ and $b=k(x+1)$, where perhaps $k$ is very large. Then $a^p-b^p=k^p\left((x+1)^p-x^p\right)$, and $(a-b)^p=k^p$, so our inequality becomes $$(x+1)^p -x^p \le C(p).$$ If $x$ is large enough, this cannot hold.

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