Áp dụng BĐT AM-GM ta có:
\(\left(a+1\right)^2+b^2+1=a^2+2a+1+b^2+1=\left(a^2+b^2\right)+2a+2\ge2\left(ab+a+1\right)\)
\(\Rightarrow\frac{1}{\left(a+1\right)^2+b^2+1}\le\frac{1}{2\left(ab+a+1\right)}\)(1)
\(\left(b+1\right)^2+c^2+1=b^2+2b+1+c^2+1=\left(b^2+c^2\right)+2b+2\ge2\left(bc+b+1\right)\)
\(\Rightarrow\frac{1}{\left(b+1\right)^2+c^2+1}\le\frac{1}{2\left(bc+b+1\right)}\)(2)
\(\left(c+1\right)^2+a^2+1=c^2+2c+1+a^2+1=\left(c^2+a^2\right)+2c+2\ge2\left(ca+c+1\right)\)
\(\Rightarrow\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2\left(ca+c+1\right)}\)(3)
Cộng vế theo vế của (1) ; (2) ; (3) ta được:
\(\frac{1}{\left(a+1\right)^2+b^2+1}+\frac{1}{\left(b+1\right)^2+c^2+1}+\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{1}{2}\)Dấu "=" xảy ra \(\Leftrightarrow a=b=b=1\)
