Is my proof correct? (standard algebra, some calculus) I'm teaching myself mathematics and I tried to solve a problem from the STEP exam (Cambridge's admission test). The solution is considerably shorter and less "clumsy" than mine but I would like to know if my proof is at least correct.
Thank you.

We have $$f'(x)=a-\frac{x^4+3x^2}{x^4+2x^2+1}$$ and we want to prove that if $a\geq\frac{9}{8}$ then $f'(x)\geq 0$ for all x. \
If the statement is proven for $a=\frac{9}{8}$, the case of $a > \frac{9}{8}$ follows. Therefore we only need to prove that $$ \frac{9}{8} \geq \frac{x^4+3x^2}{x^4+2x^2+1} $$
The proof:
 \begin{gather}
  \frac{9}{8} \geq \frac{x^4+3x^2}{x^4+2x^2+1} \\
  \frac{9(x^4+2x^2+1)-8(x^4+3x^2)}{8(x^4+2x^2+1)} \geq 0 \\
  \frac{x^4-6x^2+9}{8x^4+16x^2+8} \geq 0 \;\; \text{(3)}
 \end{gather}
Since $\forall x \; 8x^4+16x^2+8 > 0 $, in fact $ \geq 8 $, then for (3) to be incorrect $ x^4-6x^2+9 < 0 $. I will prove this is impossible for all x.
 \begin{gather*}
  x^4-6x^2+9 < 0 \\
  x^4-6x^2 < -9
 \end{gather*}
However the global minimum of $ x^4-6x^2 $ is $-9$:
$\frac{d}{dx}x^4-6x^2=4x^3-12x=0$ when $x = 0, x =\sqrt{3}$ and $x=\sqrt{3}$ ,when plugged back in, yields $ 9-6*3 = -9 $ the global minimum.
It follows that $ x^4-6x^2 \geq -9 $.
Therefore the statement (3) is correct and I have proved that, when $a \geq \frac{9}{8}$ $$ f'(x) \geq 0 $$

Yes you have done it in the right sense .
