I was trying to prove the following inequality:

Take an arbitrary $p\in\mathbb{N}$, then


This looks an awful lot like young's inequality to me, but I couldn't quite figure out how to find an analogy between the two that could help.

I've also considered proving this via induction, but for $b\leq{}a$ it doesn't seem to work. I'd greatly appreciate any tips or suggestions.

Thanks a lot.


Dividing over $a^p$, we get an equivalent inequality $(1+x)^p \le 1+\varepsilon + C_\varepsilon x^p$. When $x\le x_\varepsilon:=(1+\varepsilon)^{1/p}$, the inequality is evident. For $x>x_\varepsilon$, it is evident as well with $C_\varepsilon = (1+x_\varepsilon^{-1})^p$.


$\varepsilon$ and a, b positifs numbers you can take

$C_\varepsilon \geq \frac{(a+b)^p-(1+\varepsilon)a^p}{b^p}$


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