Từ giả thiết ta có \(1+c^2=ab+bc+ac+c^2=\left(a+c\right)\left(b+c\right)\) ; \(1+a^2=ab+bc+ac+a^2=\left(a+b\right)\left(a+c\right)\)
\(1+b^2=ab+bc+ac+b^2=\left(b+a\right)\left(b+c\right)\)
Suy ra \(\frac{a+b}{1+c^2}+\frac{b+c}{1+a^2}+\frac{c+a}{1+b^2}=\frac{a+b}{\left(c+a\right)\left(c+b\right)}+\frac{b+c}{\left(a+b\right)\left(a+c\right)}+\frac{c+a}{\left(b+a\right)\left(b+c\right)}\)
\(=\frac{\left(a+b\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(c+a\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Theo BĐT Cauchy , ta có : \(\frac{\left(a+b\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{27\left(a+b\right)^2}{\left(a+b+b+c+c+a\right)^3}=\frac{27\left(a+b\right)^2}{8\left(a+b+c\right)^3}\)
Tương tự : \(\frac{\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{27\left(b+c\right)^2}{8\left(a+b+c\right)^3}\) ; \(\frac{\left(c+a\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{27\left(c+a\right)^2}{8\left(a+b+c\right)^3}\)
\(\Rightarrow\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{9}{8\left(a+b+c\right)^3}.3\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\right]\)
\(\ge\frac{9}{8\left(a+b+c\right)^3}.\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^2\) (Áp dụng BĐT Bunhiacopxki)
\(=\frac{9.4\left(a+b+c\right)^2}{8\left(a+b+c\right)^3}=\frac{9}{2\left(a+b+c\right)}\) (đpcm)
