Problem source

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

Solution source

\begin{align*}
&& f(x) &= a x-\frac{x^{3}}{1+x^{2}} \\
\Rightarrow && f'(x) &= a - \frac{3x^2(1+x^2)-x^3 \cdot 2 x}{(1+x^2)^2} \\
&&&= a - \frac{-x^4+3x^2}{(1+x^2)^2} \\
&&&= a - \frac{-t^2+3t}{(1+t)^2} \\
&&&= \frac{a+2at+at^2-t^2-3t}{(1+t)^2} \\
&&&= \frac{(a-1)t^2+(2a-3)t+a}{(1+t)^2} \\
\\
&& 0 \leq \Delta  &= (2a-3)^2 - 4 \cdot (a-1) \cdot a \\
&&&= 4a^2-12a+9 - 4a^2+4a \\
&&&= -8a + 9 \\
\Leftrightarrow && a &\geq 9/8
\end{align*}

Therefore if $a \geq 9/8$ the numerator is always non-negative and $f'(x) \geq 0$