Xét: \(\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\)
\(\Leftrightarrow a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}+1-\dfrac{a^2}{a^2+1}+1-\dfrac{b^2}{b^2+1}+1-\dfrac{c^2}{c^2+1}\)
\(\Leftrightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)+3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Xét \(3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) ( 1 )
Theo hệ quả của bất đẳng thức Cauchy ta có
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\) ( 3 )
Xét \(3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2}{a^2+1}\le\dfrac{a^2}{2a}=\dfrac{a}{2}\\\dfrac{b^2}{b^2+1}\le\dfrac{b^2}{2b}=\dfrac{b}{2}\\\dfrac{c^2}{c^2+1}\le\dfrac{c^2}{2c}=\dfrac{c}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Rightarrow3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\) ( 4 )
Từ ( 3 ) và ( 4 ) cộng theo từng vế
\(\Rightarrow VT\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)
\(\Leftrightarrow\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\ge3\)
\(\Rightarrow\) ( đpcm )
