Mình cũng xin góp thêm 1 cách giải:
\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3\sqrt[3]{1} = 3\]
Lại có: \[\left( {{a^2} + {b^2} + {c^2}} \right)\left( {{1^2} + {1^2} + {1^2}} \right) \ge {\left( {a + b + c} \right)^2}\]
Hay:\[\frac{{{{\left( {a + b + c} \right)}^2}}}{3} \le {a^2} + {b^2} + {c^2}\]
Suy ra:\[\frac{{a + b + c}}{{\sqrt {{a^2} + {b^2} + {c^2}} }} \ge \frac{{a + b + c}}{{\sqrt {\frac{{{{\left( {a + b + c} \right)}^2}}}{3}} }} = \frac{{a + b + c}}{{\frac{{a + b + c}}{{\sqrt 3 }}}} = \sqrt 3 \]
Vậy:\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{{a + b + c}}{{\sqrt {{a^2} + {b^2} + {c^2}} }} \ge 3 + \sqrt 3 \]