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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 $n \geq 2$, $0<s<n-1$ and $\gamma$ a real in (0,1).

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Under what theory are asked to prove this? –  Pedro Tamaroff Apr 30 '12 at 18:38
    
@Peter Tamaroff: Any correct proof or counter-example provided will do. –  pebox11 Apr 30 '12 at 18:42
    
I'm just saying it because the $\gamma(s-1)$ reminds me of convex functions. –  Pedro Tamaroff Apr 30 '12 at 18:52
    
@Peter Tamaroff: you can use theory of convex functions if you wish. –  pebox11 Apr 30 '12 at 19:03
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peb, the thought process in your previous two comments would be a welcome addition to the body of your question. –  The Chaz 2.0 Apr 30 '12 at 19:33

1 Answer 1

up vote 1 down vote accepted

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. –  pebox11 Apr 30 '12 at 20:50
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No, we don't. $f$ is decreasing on $(0,\infty)$. Nothing special happens as $\gamma \to 1$. –  Robert Israel Apr 30 '12 at 21:35
    
Yes you are correct, the monotonicity will solve the problem. So the solution is complete. –  pebox11 Apr 30 '12 at 21:58
    
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)... –  pebox11 May 2 '12 at 7:49
    
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)); –  Robert Israel May 2 '12 at 8:07

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