I want to show that this inequality holds for $0 . $ q^{-p}(1-q)^{p-1} \geq (1-p)^{p-1}p^{-p} $ I have tried to divide it into 3 cases: But it doesn't seem to work!
simple inequality
3
$\begingroup$
calculus
inequality
-
0Is it right:$q^{-p}$ and $p^{-p}$ ? – 2012-11-30
1 Answers
2
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} $