Câu bất đẳng thức:
$\begin{array}{l} S = \dfrac{{ab\sqrt {c - 4} + bc\sqrt {a - 5} + ca\sqrt {b - 6} }}{{abc}}\\ S = \dfrac{{\sqrt {c - 4} }}{c} + \dfrac{{\sqrt {a - 5} }}{a} + \dfrac{{\sqrt {b - 6} }}{b}\\ + )4 + \left( {c - 4} \right) \ge 2\sqrt {4\left( {c - 4} \right)} = 4\sqrt {c - 4} \Leftrightarrow \sqrt {c - 4} \le \dfrac{c}{4} \Leftrightarrow \dfrac{{\sqrt {c - 4} }}{c} \le \dfrac{1}{4}(1)\\ + )5 + \left( {a - 5} \right) \ge 2\sqrt {5\left( {a - 5} \right)} = 2\sqrt 5 .\sqrt {a - 5} \Leftrightarrow \dfrac{{\sqrt {a - 5} }}{a} \le \dfrac{1}{{2\sqrt 5 }}(2)\\ + )6 + \left( {b - 6} \right) \ge 2\sqrt {6\left( {b - 6} \right)} = 2\sqrt {6.\left( {b - 6} \right)} \Leftrightarrow \dfrac{{\sqrt {b - 6} }}{b} \le \dfrac{1}{{2\sqrt 6 }}(3)\\ (1) + \left( 2 \right) + \left( 3 \right) \Leftrightarrow S \le \dfrac{1}{4} + \dfrac{1}{{2\sqrt 5 }} + \dfrac{1}{{2\sqrt 6 }}\\ = \Leftrightarrow \left\{ \begin{array}{l} c - 4 = 4 \Leftrightarrow c = 8\\ a - 5 = 5 \Leftrightarrow a = 10\\ b - 6 = 6 \Leftrightarrow b = 12 \end{array} \right. \Rightarrow S \le \dfrac{1}{4} + \dfrac{1}{{2\sqrt 5 }} + \dfrac{1}{{2\sqrt 6 }} \Rightarrow \max S = \dfrac{1}{4} + \dfrac{1}{{2\sqrt 5 }} + \dfrac{1}{{2\sqrt 6 }} \end{array}$
