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Is there any easy way of simplifying the following inequality

$$ (a+\frac{\Delta}{2})^{\gamma} - (b-\frac{\Delta}{2})^{\gamma} > a^{\gamma} - b^{\gamma} $$

if $\gamma \in \mathbb{R}^{+}$ and $a>b>\Delta$.

Another option might be trying to find a bound for how large this difference can be.

Thanks!

econ_ugrad
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1 Answers1

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Write the question as follows, and obtain the consequent equivalent formulations: $$ (a+\frac{\Delta}{2})^{\gamma} - a^{\gamma} > (b-\frac{\Delta}{2})^{\gamma}- b^{\gamma}\\ \int_0^\Delta \frac{\gamma}{2}(a+\frac{t}{2})^{\gamma-1} \, dt> - \int_0^\Delta \frac{\gamma}{2}(b-\frac{t}{2})^{\gamma-1} \, dt\\ \frac{\gamma}{2} \int_0^\Delta (a+\frac{t}{2})^{\gamma-1} + (b-\frac{t}{2})^{\gamma-1} \, dt> 0 $$ Now since the integrand is positive, this proves the claim.$\qquad \Box$

Andreas
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  • Thanks for the very correct answer. However, I am not as interested in proving the statement. I am more interested into bounding it or simplifying it. Is there any easy bound (which can depend on $\delta,a$ and $b$ on this integral? – econ_ugrad Feb 04 '21 at 18:54
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    Very rough, you could argue: for $\gamma >1$, $(a+\frac{t}{2})^{\gamma-1} + (b-\frac{t}{2})^{\gamma-1} > a^{\gamma-1} + (b-\frac{\Delta}{2})^{\gamma-1} $, so the LHS is bounded by $\frac{\gamma \Delta}{2} [a^{\gamma-1} + (b-\frac{\Delta}{2})^{\gamma-1} ]$. For $\gamma \le 1$, with the same reasoning you have the lower bound $\frac{\gamma \Delta}{2} [ (a+\frac{\Delta}{2})^{\gamma-1} + b^{\gamma-1}]$. – Andreas Feb 04 '21 at 19:03
  • I am not sure how I follow from the first inequality how that term becomes an upper bound. Nonetheless, this seems to me to be the kind of bound I am looking for! Do you mind to elaborate a little bit in the steps? – econ_ugrad Feb 04 '21 at 20:06
  • The bounds I gave are lower bounds of the LHS. Just take the smallest value of the integrands, multiply with the width. – Andreas Feb 04 '21 at 22:12