# Decreasing function made up of cubes, squares and floor function

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. – JavaMan Jan 23 '12 at 17:03

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.
More simply: Fix $r \geq 1$ and let $t > 0$. As you say, it suffices to show that $g(t) = 2(t^2 + r)^{3/2} - (2t^3 + 3rt)$ is nonincreasing in $t$. Compute \begin{align} &g'(t) = 6t(t^2 + r)^{1/2} - (6t^2 + 3r) \leq 0 \\ \iff &(t^2 + r)^{1/2} \leq t +\frac{r}{2t} \\ \iff &t^2 + r \leq t^2 + r + \frac{r^2}{4t^2} \\ \iff &\frac{r^2}{4t^2} \geq 0 \end{align} with the last statement being clearly always true. Therefore $g(t)$ is decreasing in $t$ as desired. – JavaMan Jan 23 '12 at 22:17
@David: By the way, it took me a long time to correctly read your post. I believe what you are saying is that if $h(t) = 6t(t^2 +r)^{1/2} + 6t^2 +3r$, then since $h(t) \geq 0$ for all $r,t$ and $\frac{\partial g}{\partial t} \cdot h(t) < 0$, it must be that $\frac{\partial g}{\partial t} < 0$ for all $r,t$. Is that correct? – JavaMan Jan 23 '12 at 22:19