+ \(\frac{1}{a^2+2b^2+3}=\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\frac{1}{2\left(ab+b+1\right)}\) . Dấu "=" \(\Leftrightarrow a=b=1\)
+ Tương tự : \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\). Dấu "=" \(\Leftrightarrow b=c=1\)
\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ca+a+1\right)}\). Dấu "=" \(c=a=1\)
Do đó : \(VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{abc\cdot b+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}\right)=\frac{1}{2}\)
Dấu "=" \(\Leftrightarrow a=b=c=1\)
