\(S\ge3\sqrt[6]{\frac{a^2b^2+1}{ab}.\frac{b^2c^2+1}{bc}.\frac{c^2a^2+1}{ca}}\)
Sở trường của Thắng. ( làm rùm) mình tịt rồi.
\(\left(a^2+\frac{1}{b^2}\right)\left(1^2+4^2\right)\ge\left(a+\frac{4}{b}\right)^2\)
=> \(a^2+\frac{1}{b^2}\ge\frac{1}{17}\left(a+\frac{4}{b}\right)^2\Leftrightarrow\sqrt{a^2+\frac{1}{b^2}}\ge\frac{1}{\sqrt{17}}\left(a+\frac{4}{b}\right)\)
CM tương tự \(\sqrt{b^2+\frac{1}{c^2}}\ge\frac{1}{\sqrt{17}}\left(b+\frac{4}{c}\right);\sqrt{c^2+\frac{1}{a^2}}\ge\frac{1}{\sqrt{17}}\left(c+\frac{4}{a}\right)\)
=> \(S\ge\frac{1}{\sqrt{17}}\left(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)\)
\(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}=\left(16+\frac{4}{a}\right)+\left(16b+\frac{4}{b}\right)+\left(16c+\frac{4}{c}\right)-15\left(a+b+c\right)\)
\(\ge16+16+16-15\cdot\frac{3}{2}=\frac{51}{2}\)
=> S >= ...
Vậy GTNN của S là ... tại a = b= c = 1/2
