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Let $a>b > 0$ and $0 < \delta < 1$, then does there exist a $c$ such that \begin{align*} \left( \dfrac{a}{b} \right)^{\delta} & \leq \left( \dfrac{a+1}{b+1} \right)^{\delta + c}\\ \end{align*}

Of course if $c = 0$, we get the opposite inequality due to this $$ \left( \dfrac{a}{b} \right)^{\delta} > \left( \dfrac{a+1}{b+1} \right)^{\delta } \,. $$

It also seems that $c$ must be more than or equal to 1, but is there a generic construction of $c$?

Greenparker
  • 542
  • Playing with logarithm and the fact that $\ln(x+1) = \ln(x) + \ln(1+1/x)$ we can get : $$ \begin{align} \delta(\ln a - \ln b) \leq (\delta + c) (\ln(a) + \ln(1+1/a) - \ln(b) - \ln(1+1/b) ) \Leftrightarrow \ 0 \leq \delta(\ln(1+1/a) - \ln(1+1/b)) + c( \ln(a+1) - \ln(b+1) ) \Leftrightarrow \ -\frac{\delta(\ln(1+1/a) - \ln(1+1/b))}{ \ln(a+1) - \ln(b+1) } \leq c \end{align} $$ – Zubzub Apr 24 '17 at 09:21
  • Thanks @Zubzub. That's helpful, but I would like $c$ to be independent of $b$. I know this might not be possible without knowing more about $b$, but I was curious if the community had any ideas. – Greenparker Apr 24 '17 at 09:30

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