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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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    Under what theory are asked to prove this?2012-04-30
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    @Peter Tamaroff: Any correct proof or counter-example provided will do.2012-04-30
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    I'm just saying it because the $\gamma(s-1)$ reminds me of convex functions.2012-04-30
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    @Peter Tamaroff: you can use theory of convex functions if you wish.2012-04-30
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    Say you try to bound the left hand side from above: $\frac{2}{\gamma}[\sqrt{(1+\gamma (n-1))(1+\gamma (s -1))}-(1+\gamma (s -1))] \leq \frac{2}{\gamma}[\sqrt{(1+\gamma (n-1))^2}-(1+\gamma (s -1))] \leq \frac{2}{\gamma}\gamma(n-s)$ which is not a good bound2012-04-30
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    Say now that you would like to see what happens when $\gamma$ approaches $0$, you will then have a situation in which $\frac{2}{0}0$ in which you can not say anything.2012-04-30
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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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    Yes, this is correct as $\gamma$ approaches $0$ from the right. But it is the half way there. We have to check also what happens as $\gamma$ tends to 1.2012-04-30
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    No, we don't. $f$ is decreasing on $(0,\infty)$. Nothing special happens as $\gamma \to 1$.2012-04-30
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    Yes you are correct, the monotonicity will solve the problem. So the solution is complete.2012-04-30
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    Is there a possibility that in the numerator of the expression of $f^'$ you provided, instead of -2 we should have -4? I just wanted to verify the calculations again (using both maple and paper) and it seems this way... This -4 might make some difference in applying L'Hospital (have not checked this yet)...2012-05-02
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    This is exactly what Maple gave me, and in this instance I trust it. Here's my Maple code (note that I use $g$ instead of $\gamma$, which to Maple is a constant). > f:= 2/g*(sqrt((1+g*(n-1))*(1+g*(s-1))) - (1+g*(s-1))); > simplify(diff(f, g));2012-05-02
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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