Dùng BĐT Cauchy
[tex]a+\frac{1}{b}=a+9.\frac{1}{9b}=a+\frac{1}{9b}+\frac{1}{9b}+...+\frac{1}{9b} \geq 10\sqrt[10]{\frac{a}{9^{9}.b^{9}}}[/tex] (9 số [tex]\frac{1}{9b}[/tex] ) (1)
Tương tự có:
[tex]b+9.\frac{1}{9c} \geq 10\sqrt[10]{\frac{b}{9^{9}.c^{9}}}[/tex] (2)
[tex]c+9.\frac{1}{9a} \geq 10\sqrt[10]{\frac{c}{9^{9}.a^{9}}}[/tex] (3)
Từ (1) (2) (3) có:
[tex](a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{a}) \geq 10^{3}.\sqrt[10]{\frac{1}{9^{27}a^{8}b^{8}c^{8}}}[/tex]
Dùn BĐT [tex]abc\leq (\frac{a+b+c}{3})^3[/tex] có:
[tex]a^{8}b^{8}c^{8} \leq (\frac{a+b+c}{3})^{24}=\frac{1}{9^{12}}[/tex]
Vầy [tex](a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{a}) \geq 10^{3}\sqrt[10]{\frac{1}{9^{15}.\frac{1}{9^{12}}}}=(\frac{10}{3})^{3}[/tex]