Young's inequality or something similar

Question

I was trying to prove the following inequality:

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

$\forall\varepsilon>0\,\forall{}a,b\in\mathbb{R}_{+}\exists{}C_{\varepsilon}>0\,:\,(a+b)^p\leq{}(1+\varepsilon)a^p\,+\,C_{\varepsilon}b^p$

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.

Answer 1

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 $C_\varepsilon = (1+x_\varepsilon^{-1})^p$.

Answer 2

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

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