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Let $t$ and $r$ be two integers with $r\geq 1, t\geq \frac{r}{2}$. Put $ f(r,t)=\lfloor 2(t^2+r)^{\frac{3}{2}}-(2t^3+3rt) \rfloor $

(here $\lfloor x \rfloor$ denotes the floor of $x$, i.e. the largest integer below $x$). Thus $f(1,.)$ is identically zero, the first two values of $f(2,.)$ are $2$ and $1$, followed by zeroes. It is easy to see that $f(r,t)=0$ whenever $t\geq \frac{3}{4}r^2$. Is it true that $f(r,.)$ is decreasing (as a function of $t$) for every $r \geq 1$ ?

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    (Just an idea:) One way to get ahold ofthis problem is to fix $r$, and let $g(t) = 2(t^2+r)^{3/2} - (2t^3 + 3rt)$. Compute $g'(t)$ to see that _eventually_ $g'(t) \leq 0$ for $t$ sufficiently large.2012-01-23

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Assume that $t\ge 0$ and $r>0$, and set $g(r,t):=2(t^2+r)^{3/2}-(2t^3+3rt)$. Then $\partial g/\partial t=6t\sqrt{t^2+r}-(6t^2+3r)$. Now if we multiply by the conjugate $ (6t\sqrt{t^2+r}+6t^2+3r)(\partial g/\partial t)=-9r^2< 0, $ so we also have $\partial g/\partial t< 0$. Therefore, $g(r,t)$ is nonincreasing in $t$, so $f(r,t)=\lfloor g(r,t) \rfloor$ must be as well.

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    Re the comment by @JavaMan immediately above, that's correct.2012-01-23