Cho các số thực dương \(a,\,\,b,\,\,c\) thỏa mãn \(ab + bc + ca = abc.\)

1)Ta có \(ab + bc + ca = abc \Leftrightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\).

Cách 1: Khi đó \(a + b + c = (a + b + c)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \ge 3\sqrt[3]{{abc}}.3\sqrt[3]{{\frac{1}{{abc}}}} = 9\)

Cách 2: \(a + b + c = (a + b + c)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) = 3 + \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{c}{a} \ge 3 + 2 + 2 + 2 = 9\)

2)Ta có \(a + b + c = (a + b + c)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) = \frac{{{{\left( {a + b} \right)}^2}}}{{ab}} + \frac{{{{\left( {b + c} \right)}^2}}}{{bc}} + \frac{{{{\left( {c + a} \right)}^2}}}{{ca}} - 3\)

\( \ge \frac{{4{{\left( {a + b + c} \right)}^2}}}{{ab + bc + ca}} - 3 = \frac{{4\left( {{a^2} + {b^2} + {c^2}} \right)}}{{abc}} + 5 = 4\left( {\frac{a}{{bc}} + \frac{b}{{ca}} + \frac{c}{{ab}}} \right) + 5\)