Sử dụng bất đẳng thức quen thuộc: \(4ab\le\left(a+b\right)^2\)
Ta có:
\(\Rightarrow\left[\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\right]^2\le\frac{1}{4}\)
\(\Rightarrow\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\)
Vậy \(\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\left(đpcm\right)\)