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.