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I want to show that this inequality holds for $0<p,q<1$.

$ q^{-p}(1-q)^{p-1} \geq (1-p)^{p-1}p^{-p} $

I have tried to divide it into 3 cases:

  1. $0<p<q<1$
  2. $0<q<p$
  3. $p=q$ (obvious)

But it doesn't seem to work!

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Is it right:$q^{-p}$ and $p^{-p}$ ? – HipsterMathematician Nov 30 '12 at 21:20
up vote 2 down vote accepted

Consider $$ f(x)=-p\log(x)+(p-1)\log(1-x)\tag{1} $$ Then $$ \begin{align} f'(x)&=-\frac{p}{x}-\frac{p-1}{1-x}\\ f''(x)&=\frac{p}{x^2}+\frac{1-p}{(1-x)^2} \end{align}\tag{2} $$ $f'(x)=0$ when $$ \frac{p}{x}=\frac{1-p}{1-x}\tag{3} $$ One solution to $(3)$ is when $x=p$, and since $f''(x)\gt0$ for all $x$, this is a unique minimum.

Therefore, for any $q$, we must have $$ f(q)\ge f(p)\tag{4} $$ which implies $$ q^{-p}(1-q)^{p-1}\ge p^{-p}(1-p)^{p-1}\tag{5} $$

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