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I have a quite complicated function of 2 variables:

```
f[n_, dj_] =
(6*(2 - n + dj*(5 + 2*n))*(7 + n + dj*(5 + 2*n))*(-965 +
2*dj^4*(5 + 2*n)^4 + dj^3*(5 + 2*n)^3*(35 + 2*n) -
9*dj^2*(5 + 2*n)^2*(-19 + n + 2*n^2) +
dj*(-1 + 2*n)*(5 + 2*n)*(-73 + n*(-26 + 11*n)) -
n*(473 + n*(111 + n*(-37 + 8*n)))))/((5 - n +
dj*(5 + 2*n))*(1 - 5*n + dj*(33 + 6*n + 4*dj*(5 + 2*n)))*
Sqrt[(5 + 2*n)^2*(5 - n + dj*(5 + 2*n))*(1 - 5*n +
dj*(33 + 6*n + 4*dj*(5 + 2*n)))*(2477 + 12*dj^4*(5 + 2*n)^4 +
4*dj^3*(5 + 2*n)^3*(59 + 2*n) -
dj^2*(5 + 2*n)^2*(-1573 + 4*n*(1 + 5*n)) +
n*(-2230 + n*(-411 + 4*n*(29 + 8*n))) -
2*dj*(5 + 2*n)*(-1937 + n*(435 + 2*n*(87 + 8*n))))])
```

I want to show that this function is always less than 1 for $n \geq d_j \geq 1$. Indeed when I do

```
Resolve[ForAll[{n, dj}, n >= dj >= 1, f[n, dj] < 1]]
```

I get `True`

. I'm just wondering if there's a way I could show this more rigorously (i.e. "by hand," or ask *Mathematica* to show steps it used to get `True`

), since I need to put this in a paper.

Duplicate: http://mathematica.stackexchange.com/q/148/5

– rm -rf – 2013-03-15T20:21:19.9301Are you genuinely interested in the mathematical details or are you looking for some additional reassurance of correctness? In the latter case, perhaps a plot would be useful. – Mark McClure – 2013-03-15T20:31:03.077

I guess the former case. I really believe it's true. Just not sure if just showing that Mathematica says so would convince other people (for publication, for example). – Art – 2013-03-15T20:41:24.620

Quoting from the answer rm linked to, "how Mathematica does things doesn't necessarily correspond to "manual" methods". Even if you knew the steps Mathematica does internally, in the majority of cases this would not be useful at all. What I'd do for publishing this is that I'd verify the result numerically and mention it that it agrees with what Mathematica told me. I would not trust Mathematica alone on this, but a numerical verification would convince me. – Szabolcs – 2013-03-15T21:21:05.893

If you are looking for a computed-aided proof, I would suggest looking into the subject of "quantifier elimination." – James Rohal – 2013-03-15T21:30:20.383

(But of course only if it's not possible to prove it in an easy way otherwise. An easy proof would only be possible if we knew how you got the expression.) – Szabolcs – 2013-03-15T21:38:56.847

Have you looked up

`Tracescan`

? – Jonathan Shock – 2013-03-15T22:28:48.343