Prove a well-known inequality using Cauchy-Schwarz or AM-GM For $a;b>0$ and $ab \geq 1$ we have a well-known inequality:
$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2} \geq \dfrac{2}{1+ab}$
Which is equivalent to: $\dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)} \geq 0$ (true)
But is there a solution for this inequality by Cauchy-Schwarz or AM-GM? Thank you.

Again not AM-GM or C-S (unless you count $a^2+b^2\geq2ab$) but I thought this was kinda sweet:
\begin{align*}
\frac{1}{1+a^2}+\frac{1}{1+b^2}=1-\frac{(ab)^2-1}{1+a^2+b^2+(ab)^2}&\geq1-\frac{(ab)^2-1}{1+2ab+(ab)^2}\\
&=1-\frac{ab-1}{ab+1}\\
&=\frac{2}{1+ab}.
\end{align*}
