Ta có: \(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{\left(a+1\right)b^2}{b^2+1}\)
\(\ge\left(a+1\right)-\frac{\left(a+1\right)b^2}{2b}=a+1-\frac{ab+b}{2}\)
Tương tự ta có:\(\frac{b+1}{c^2+1}\ge b+1-\frac{bc+c}{2};\frac{c+1}{a^2+1}\ge c+1-\frac{ca+a}{2}\)
Cộng theo vế ta có: \(VT\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}=6-\frac{3+ab+bc+ca}{2}\)
Mà theo BĐT AM-GM: \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Suy ra \(VT\ge6-3=3\)(ĐPCM)