Ta có: \(\left\{\begin{matrix}a^2+b^2\ge2ab\\b^2+1\ge2b\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2ab+2b+2}=\frac{1}{2\left(ab+b+1\right)}\)
Tương tự ta có:\(\left\{\begin{matrix}\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\\\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ac+a+1\right)}\end{matrix}\right.\)
Cộng theo vế của 3 BĐT trên ta có:
\(VT\le\frac{1}{2\left(ab+b+1\right)}+\frac{1}{2\left(bc+c+1\right)}+\frac{1}{2\left(ac+a+1\right)}\)
\(=\frac{1}{2}\left(\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}\right)\left(abc=1\right)\)
\(=\frac{1}{2}\left(\frac{ac+a+1}{ac+a+1}\right)=\frac{1}{2}\) (Đpcm)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}abc=1\\a=b=c\\a,b,c>0\end{matrix}\right.\)\(\Rightarrow a=b=c=1\)