Ta có: \(\frac{a}{1+b^2}=\frac{a+ab^2-ab^2}{1+b^2}=\frac{a+ab^2}{1+b^2}-\frac{ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự ta có: \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\); \(\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Cộng theo vế 3 BĐT trên,ta được: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{1}{2}\left(ab+bc+ca\right)\)
Do \(ab+bc+ca\ge\frac{\left(a+b+c\right)^2}{3}\) (dấu "=" xảy ra khi a = b = c) nên ta có:)
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{1}{2}\left(ab+bc+ca\right)\ge3-\frac{1}{2}.\frac{\left(a+b+c\right)^2}{3}=\frac{3}{2}^{\left(đpcm\right)}\)