proving one function is “greater” than another?

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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Hmm, we want $$\alpha(l-f)^{\alpha - 1} > \frac{(l-f)^\alpha}{l}$$Since $l$ and $\frac{1}{(l-f)^{\alpha - 1}}$ are both both positive, we can multiply both sides of the inequality by $\frac{l}{(l-f)^{\alpha-1}}$ and not change its sign, we arrive at: $$\alpha l > (l - f)$$ This seems false...namely, divide both sides by $l$, and consider the limit as $l \rightarrow \infty$, we get $$\alpha > 1$$or am I missing something :(? – uncookedfalcon Apr 9 '12 at 6:13
Seems true iff $f > l (1-\alpha)$. – copper.hat Apr 9 '12 at 6:26
ah, so it is not true for all $\alpha\in (0,1)$? – Emir Apr 10 '12 at 7:48
@uncookedfalcon You get $\alpha\ge 1$ in the limit, not $\alpha>1$. – ˈjuː.zɚ79365 Jun 15 '13 at 0:15

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 $f>l(1-\alpha)$.