Given that $a \gt b\ge 1$,show that $f(b)\gt f(a)$

Question

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.

Answer 1

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 $[1,\infty)$