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Show that for $\alpha\in (0,1)$, $l,f\in\mathbb{R}_+ (l>f)$, $x(l)>y(l),\forall l\in\mathbb{R_+}$ where $y(l)=(l-f)^{\alpha}/l$ and $x(l)=\alpha(l-f)^{\alpha-1}$.

I first noted that $(l-f)>0,\forall l,f$. Then I looked at cases:
(1) for $l-f<1$, $(l-f)^{\alpha}<(l-f)^{\alpha-1}$, but then since there's the case that $l<1, (l-f)^{\alpha}/l>(l-f)^{\alpha}$...

I'm confused as to how to approach this in a rigorous way rather than just plugging in values.

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    @uncookedfalcon You get $\alpha\ge 1$ in the limit, not \alpha>1.2013-06-15

1 Answers 1

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As observed by uncookedfalcon and copper.hat in the comments, the desired inequality $x>y$ simplifies to $\alpha l > l-f$. The latter does not hold as generally as stated in the question, but it holds if we additionally assume $f>l(1-\alpha)$.