Thấy : \(\dfrac{1}{b^2+1}=1-\dfrac{b^2}{b^2+1}\ge1-\dfrac{b^2}{2b}=1-\dfrac{b}{2}\) ( AD BĐT Cô-si )
Suy ra : \(\dfrac{a+1}{b^2+1}\ge\left(a+1\right)\left(1-\dfrac{b}{2}\right)=a+1-\dfrac{ab}{2}-\dfrac{b}{2}\)
CMTT : \(\dfrac{b+1}{c^2+1}\ge b+1-\dfrac{bc}{2}-\dfrac{c}{2}\) ; \(\dfrac{c+1}{a^2+1}\ge c+1-\dfrac{ac}{2}-\dfrac{a}{2}\)
Suy ra : \(VT\ge\dfrac{a+b+c}{2}+3-\dfrac{ab+bc+ac}{2}=\dfrac{9}{2}-\dfrac{ab+bc+ac}{2}\) (1)
Mặt khác : \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\) (2)
(1) ; (2) Suy ra : \(VT\ge\dfrac{9}{2}-\dfrac{3}{2}=3=a+b+c\)
" = " \(\Leftrightarrow a=b=c=1\)
- Sửa đề: \(a+b+c\le3\)
- Ta có: \(\dfrac{a+1}{b^2+1}=a+1-\dfrac{\left(a+1\right)b^2}{b^2+1}\ge a+1-\dfrac{\left(a+1\right)b^2}{2b}=a+1-\dfrac{ab+b}{2}\left(1\right)\)
- Tương tự: \(\dfrac{b+1}{c^2+1}=b+1-\dfrac{bc+c}{2}\left(2\right)\); \(\dfrac{c+1}{a^2+1}=c+1-\dfrac{ca+a}{2}\left(3\right)\)
- Từ (1), (2), (3) suy ra:
\(\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\ge\dfrac{a+b+c}{2}+3-\dfrac{ab+bc+ca}{2}\)
Mà ta có BĐT phụ: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\ge\dfrac{a+b+c}{2}+3-\dfrac{\left(a+b+c\right)^2}{6}\ge\dfrac{a+b+c}{2}+3-\dfrac{3.\left(a+b+c\right)}{6}=\dfrac{a+b+c}{2}-\dfrac{a+b+c}{2}+3=3\ge a+b+c\)
- Dấu "=" xảy ra khi \(a=b=c=1\)
