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Is it true that for $a,b \in \mathbb{R}$ and $p\geq 1$ (or $p\geq 2)$ there exists a constant $C>0$ independent of $a,b$ of course, such that:

$(a-b)^{p-2} ab \leq C (a-b)^p$

Thanks a lot! :)

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3 Answers 3

No.

It it were true, this would hold in particular for all $b$ and all $a=b+1$.

But in this case, this would imply: $$ b(b+1)\leq C $$ for all $b$, which is clearly wrong (let $b$ tend to $+\infty$).

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Thank you very much! –  Charlie Feb 1 '13 at 14:15
    
@Charlie You're welcome. No if you want, you may upvote the answers you like and accept your prefered one. –  1015 Feb 1 '13 at 14:27

Not in general. Suppose $p\geq 1$ and $b=a+\varepsilon$, where $\varepsilon>0$ will be chosen later. Then your inequality becomes $$(2C+1)ab\leq C(a^2+b^2)?\tag{1}$$ Substitute $b=a+\varepsilon$ in $(1)$ to get $$2Ca^2+2Ca\varepsilon+a^2+a\varepsilon\leq 2Ca\varepsilon+2Ca^2+\varepsilon^2\tag{2}$$ which simplifies into $$a^2+a\varepsilon\leq C\varepsilon^2\tag{3}$$ which is absurd taking $a$ large and $\varepsilon\to 0$.

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I think it is untrue.

$$ C \geq \frac{ (a - b)^{p} \cdot a b }{ (a - b)^{p + 2}} $$ where $a \neq b$.

or

$$ C \ge \frac{ab}{(a - b)^2} = \frac{ab}{a^2 + b^2 - 2ab}$$ which is easy to prove existent dependent on values of $a$ and $b$.

Alternatively, just use 2 different set of input values. If $C$ is not constant, then it is dependent on either $a$ or $b$ or both.

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No, it isn't true. –  1015 Feb 1 '13 at 14:15
    
@julien untrue it was meant to be. –  hjpotter92 Feb 1 '13 at 14:16
    
Oh? well if $a\sim b$ it blows up unfortunately :( –  Charlie Feb 1 '13 at 14:17

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