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Ta có:${\left( {a + b - c} \right)^2} \ge 0$
$ \leftrightarrow {a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ac \ge 0$
$ \leftrightarrow {a^2} + {b^2} + {c^2} \ge 2\left( {bc + ac - ab} \right)$
$ \leftrightarrow 2\left( {bc + ac - ab} \right) \le \dfrac{5}{3} < 2$
$ \leftrightarrow bc + ac - ab < 1$
$ \leftrightarrow \dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c} < \dfrac{1}{{abc}}$
 
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