Given that $a \gt b\ge 1$,show that $f(b)\gt f(a)$ Given that:
$$f(x)=\frac{x}{x^{2}+1}$$
If $a \gt b \ge 1$ and both are real numbers,show that $f(b)\gt f(a)$.
My attempt:
\begin{equation}\begin{aligned} f(b)-f(a) 
&=\dfrac{b(a^2+1)-a(b^2+1)}{(b^2+1)(a^2+1)} \\
&=\dfrac{a^2b-ab^2+b-a}{(b^2+1)(a^2+1)} \\
&=\dfrac{ab(a-b)-(a-b)}{(b^2+1)(a^2+1)} \\
&=\dfrac{(ab-1)(a-b)}{(b^2+1)(a^2+1)}\\
\end{aligned}\end{equation}
Both $b^{2}+1$ and $a^{2}+1$ are $\gt 1$
And $(a-b)\gt 0$
As $ab\gt 1$,then $ab-1\gt 0$ 
Since all the factors of$f(b)-f(a)$ are positive and not zero,
$f(b)- f(a)\gt 0$
$f(b) \gt f(a)$
Is my attempt correct? If not, is there are any mistakes? If so,please check for me and show it to me. Thanks.

Your way is completely correct.
Another way to see it is just to calculate the derivative


*

*$f'(x) = \frac{1-x^2}{(x^2+1)^2} < 0$ for $x>1 \Rightarrow f$ is strictly decreasing on $[1,\infty)$

