Is this a correct way to prove this? I've just looked at this question and sketched a way to do it my head. When I looked at the answer it looked slightly more complicated than the way I did it so I just wanted to check whether this is a correct way:
The question states: 

Let $$f(x)=ax-\frac{x^3}{1+x^2}$$ where $a$ is a constant, show that
  if $a\geq9/8$ then $f'(x)\geq0$ for all $x$.

My way of doing this is to first differentiate $f(x)$ so that:
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
thus if you assume that $a\geq9/8$ and since we know from above that $$a=f'(x)+\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
then 
$$ f'(x)+\frac{x^2 (x^2+3)}{(x^2+1)^2}\geq \frac{9}{8}\;\; \rightarrow\;\;f'(x)\geq\frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2}  $$
then all we need to show is that:
$$ \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2}\geq0 $$
which we can show since we can write this as:
$$\frac{1}{8}\left( \frac{x^2-3}{x^2+1} \right)^2\geq 0$$
which is obviously true since $x^2$ is greater than or equal to zero for all real $x$.

Is this a correct way of showing this or have I inadvertently assumed something (I'm not entirely sure it's correct when I 'assume $a\geq9/8$') or made done something incorrectly? Thanks

Not very clear. If I were you I will show in this order:
first, 
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
Secondly, 
$$ \forall x, \quad \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2} = \frac{1}{8}\left( \frac{x^2-3}{x^2+1} \right)^2 \geq 0 $$
Thus if $a \geq \frac98$, then $\forall x$ we have
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2} \geq \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2} \geq 0$$
