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
Inequality holds?
1
$\begingroup$
calculus
real-analysis
analysis
functions
inequality
-
0sorry 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? – 2012-04-27
1 Answers
1
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.
-
0yes your prove is correct and so nice. Thank you very much. – 2012-04-27