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I've been trying for 2 days to prove this inequality, and I've run out of ideas. How to do it?

$$\frac{(f+c)^{f+c}}{f^f}>\frac{(1-f)^{1-f}}{(1-f-c)^{1-f-c}} , 1/2<f<1, 0<c<1-f$$

So far I've tried endless combinations of power manipulation to see if I could cancel some things which i know are bigger than 1, for instance. But I always end up with something left that I can't prove to be bigger or smaller than 1.

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    $\begingroup$ The inequality can be written as $\phi(x) > \phi(y)$ where $\phi(t)=t^{t}(1-t)^{1-t}$, $x=f+c$ and $y=f$. It looks like $\phi(t)$ is increasing for $\frac{1}{2} < t < 1$. $\endgroup$ – preferred_anon Jun 6 '18 at 7:44
  • $\begingroup$ I suppose you intended $\frac12< f< 1$ in your conditions. With the condition written as is, there is nothing to prove ;). $\endgroup$ – Macavity Jun 6 '18 at 11:16
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Hint

As $x\log x$ is convex for $x> 0$, we may use Karamata's Inequality and just show $(f+c, 1-f-c) \succ (f, 1-f)$, which is true for the conditions which imply $\frac12< f< c+f< 1$.

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  • $\begingroup$ I was looking for an algebraic proof, but this is definetly a proof, thanks. Is there any way to prove it is "greater" and not "equal or greater" than, since c is never zero? Also, c is not greater than f, but the proof holds. $\endgroup$ – Eduard Jun 6 '18 at 16:30
  • $\begingroup$ As $x \log x $ is strictly convex and $(f+c, 1-f-c) \neq (f, 1-f)$, the inequality is strict as well. $\endgroup$ – Macavity Jun 6 '18 at 17:03
  • $\begingroup$ Also you're right, we do not need $c> f$, just $f < f+c$ is enough for the majorization. Will make that edit. $\endgroup$ – Macavity Jun 6 '18 at 17:06

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