Show that , if $a \ge 9/8$, then $f '(x) \ge 0$ My questions are to do with the solution of the following question:


  Let $f(x)=ax-\frac{x^3}{1+x^2}$ , where $a$ is a positive constant.
  Show that if $a\ge\frac{9}{8}$ then $f'(x)\ge0$ for all $x$.


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

Next the following is said/done:
"We can use the inequality $a\ge\frac{9}{8}$ immediately after finding f'(x) since $a$ appears in f'(x) with a plus sign always:
$$f'(x)=\frac{a+(2a-3)x^2+(a-1)x^4}{(1+x^2)^2}\ge\frac{\frac{9}{8}+(\frac{9}{4}-3)x^2+(\frac{9}{8}-1)x^4}{(1+x^2)^2}=\frac{(x^2-3)^2}{8(1+x^2)^2}\ge0$$

My questions pertain to everything in bold and after the bold statement:


How is the author using the inequality $a\ge\frac{9}{8}$ to get to the final result?
What is the significance of $a$ appearing with a positive sign always?
First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial:
$$p(u)=(a-1)u^2+(2a-3)u+a.$$
Supposing $a\ne 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive:
$$\Delta=(2a-3)^2-4a(a-1)=9-8a\le 0.$$
So the condition is indeed $a\ge \frac98$.
Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.
We must show $(a-1)x^4+(2a-3)x^2+a\geq 0$ ,
and can easily assume $a\geq 1$. 

1: exception case If  $a=1$, $f'(x)=-x^2+1$ 

This eventually become negative. 

2: If axis$\leq0⇔ -\frac{2a-3}{a-1}\leq0⇔a\geq\frac{3}{2}$ or $a&lt;1 $, this need only $f(0)=a\geq0$ 

3: If axis $&gt;0⇔ 1&lt;a&lt;\frac32$, this need $D\leq0$ 

$D=(2a-3)^2-4a(a-1)=-8a+9\leq0$ 

We get $a\geq\dfrac98$ by these 3 cases.