If $f(x) = ax - \frac{x^3}{1+x^2}$ show that $\forall x[f'(x)\ge 0$ iff $a \ge \frac{9}{8}]$

Question

Question:

If $f(x) = ax - \frac{x^3}{1+x^2}$ show that $\forall x[f'(x)\ge 0$ iff $a \ge \frac{9}{8}]$

Part of answer found online:

$$f'(x) = a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+3x^2)^2} = a+ \frac{x^4 - 3x^2}{(1+x^2)^2}$$ My problem:

I don't understand why: $$a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+3x^2)^2} = a+ \frac{x^4 - 3x^2}{(1+x^2)^2}$$ When i try to rearrange the left i get: $$a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+3x^2)^2} = a-\frac{3x^4+3x^2-2x^4}{(1+3x^2)^2}= a - \frac{x^4+3x^2}{(1+3x^2)^2}$$ And from there: $$a - \frac{x^4+3x^2}{(1+3x^2)^2} = a + \frac{-1}{1} \cdot \frac{x^4+3x^2}{(1+3x^2)^2} = a + \frac{-x^4-3x^2}{(1+3x^2)^2}$$ Am i making a simple mistake in the interpretation of the minus in front of a fraction?

Answer 1

It's not. There is a mistake in the denominator. It should be $(1+x^2)^2$ not $(1+3x^2)^2$.

$$f^{'}(x) = a - \frac{x^4+3x^2}{(1+x^2)^2}$$

Answer 2

It should be $f'(x)=a-\frac{x^4+3x^2}{(x^2+1)^2}$.

We need to prove that $$\frac{9}{8}-\frac{x^4+3x^2}{(x^2+1)^2}\geq0,$$ which is $(x^2-3)^2\geq0$.