Prove that $a_n=na/(1+n^2 b)$ is strictly decreasing given a, b >0
I have tried proving this directly, by setting
$\frac{na}{1+n^2b} > \frac{(n+1)a}{1+(n+1)^2b}$
and end up with
$\frac{na+n^3ab+2n^2ab+nab}{(1+n^b)(1+n^2b+2bn+b)} > \frac{na+a+n^3ab+n^2ab}{(1+n^b)(1+n^2b+2bn+b)}$
I simplify that down to...
$\frac{n^2b+nb}{(1+n^2b)(1+n^2b+2bn+b)} > \frac{1}{(1+n^2b)(1+n^2b+2bn+b)} $
But I have no particular reason to believe that the inequality is true.  Am I making an algebraic error?  Is there a better way to do this?  Is there a way to do this by induction?
Thanks!

Just rewrite it as
$$
a_n = \frac{a}{\frac{1}{n} + nb}.
$$
Since $\frac{1}{n} + nb$ is increasing for sufficiently large $n$ and for $b > 0$, it shows $a_n$ must be decreasing for sufficiently large $n$ and for $a > 0$.
To be rigorous we need to say precisely what sufficiently large means and prove it. By sufficiently large, we mean that there is a $B$ such that $n > B$ implies $\frac{1}{n + 1} + (n + 1)b > \frac{1}{n} + nb$. But this is equivalent to $(n + 1)b - nb > \frac{1}{n} - \frac{1}{n + 1} \iff b > \frac{1}{n(n + 1)}$. But this condition is fulfilled if we have the stronger condition $b > \frac{1}{n^2} \implies n > \frac{1}{\sqrt{b}}$. Hence a $B$ that works is $B = \frac{1}{\sqrt{b}}$.
