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Can anyone prove that $$ \frac{\sum\limits_{i=1}^{k*} a_i i (x+\epsilon)^{(i-1)}}{\sum\limits_{i=1}^{k*} a_i (x+\epsilon)^{i}+k^*-1}>\frac{\sum\limits_{i=1}^{k*} a_i i x^{(i-1)}}{\sum\limits_{i=1}^{k*} a_i x^{i}+k^*-1} $$ where $a_i$ are some positive coefficients, $\epsilon$ is a positive number and $0<x<1$?

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Why do you think this is true? Is there any evidence? –  Sunni Apr 27 '12 at 17:49
    
False in general. For example when $k^*=1$. –  GEdgar Apr 27 '12 at 17:50
    
is that fctn $\frac 1 x$ when k* = 1 ? isn' t it positive near 0 and 0 at $\infty$ ? –  mike Apr 27 '12 at 17:52
    
sorry the sums should go from $1$ to $k^*-1$ not to $k^*$. I wrote the question from another computer. @Davide Giraudo yes $g$ is a decreasing function according to your definition and left side should be also decreasing but according to this, how can we conclude that left side is greater than the right side? –  user30124 Apr 27 '12 at 20:06
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up vote 1 down vote accepted

I think it's the reversed inequality which is true: let $f(x):=\sum_{i=1}a_ix^i$, and define $g(x)=\frac{f'(x)}{f(x)+k^*-1}=\frac{d}{dx}(\ln f(x))$. Then $g'(x)=\frac{d^2}{dx^2}(\ln f(x))$, and since $f$ is increasing, and $\log$ is concave, $x\mapsto \ln f(x)$ is concave, so $g'\leq 0$ and $g$ is decreasing.

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yes your prove is correct and so nice. Thank you very much. –  user30124 Apr 27 '12 at 23:07
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