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Prove or give a counter-example for the following:

$\frac{2}{\gamma}[\sqrt{(1+\gamma (n-1))(1+\gamma (s -1))}-(1+\gamma (s -1))] \leq n-s$

where $n,s$ natural numbers with $n \geq 2$, $0 and $\gamma$ a real in (0,1).

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    peb, the thought process in your previous two comments would be a welcome addition to the body of your question.2012-04-30

1 Answers 1

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If $f(\gamma)$ is your left side, $ \frac{df}{d\gamma} = {\frac {-2-\gamma\,s+2\,\gamma-\gamma\,n+2\,\sqrt { \left( 1+\gamma\,n -\gamma \right) \left( 1+\gamma\,s-\gamma \right) }}{{\gamma}^{2} \sqrt { \left( 1+\gamma\,n-\gamma \right) \left( 1+\gamma\,s-\gamma \right) }}}$ The limit of this as $\gamma \to 0+$ is $-(n-s)^2/4$, which is negative. If we set the numerator equal $0$, subtract the square root term from both sides, square and simplify we find $\gamma^2 (n-s)^2 = 0$. So $f'(\gamma) < 0$ on $(0,\infty)$, and in particular $f(\gamma) < \lim_{\gamma \to 0+} f(\gamma) = n-s$.

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    you are right. By mistake I was working with >f := 2*((1+(1+x*(n-1))*(1+x*(s-1)))^(1/2)-1-x*(s-1))/x in maple (and in paper) which has an extra 1 in the sqrt that introduces the extra -2 in the differentiation. Again you are very right! Thanks!2012-05-02