Áp dụng bđt Cauchy:
\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự:
\(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ac}{2}\)
Cộng theo vế: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{1}{2}\left(ab+bc+ac\right)\ge3-\frac{1}{6}\left(a+b+c\right)^2=3-\frac{3}{2}=\frac{3}{2}\)\("="\Leftrightarrow a=b=c=1\)