$f(x)=cx+\dfrac{1}{x^2+3}$. Find the values of c for which f(x) is increasing for all $x$ $f(x)=cx+\dfrac{1}{x^2+3}$. Find the values of c for which f(x) is increasing for all $x$
I found $$f'(x) =c-\dfrac{2x}{\left( x^2+3\right)^2}$$
But I am having a tough time moving from here.
I know we have to find $c$ for which the following Inequality is satisfied for all $x$
$$c-\dfrac{2x}{\left( x^2+3\right)^2}>0$$

Hint.  You need
$$c-\dfrac{2x}{\left( x^2+3\right)^2}\ge0$$
for all $x$, that is, $c$ is greater than or equal to the maximum value of $2x/(x^2+3)^2$.  Can you take it from here?
