Áp dụng $AM-GM$ cho $2$ số ta được
$\begin{array}{l} \sqrt {5{a^2} + 38ab + 21{b^2}} = \sqrt {\left( {5a + 3b} \right)\left( {a + 7b} \right)} \\ \le \dfrac{{5a + 3b + a + 7b}}{2} = \dfrac{{6a + 10b}}{2} = 3a + 5b\\ \text{Tương tự} :\sqrt {5{b^2} + 38bc + 21{c^2}} \le 3b + 5c\\ \sqrt {5{c^2} + 38ac + 21{a^2}} \le 3c + 5a\\ \Rightarrow P \le 3\left( {a + b + c} \right) + 5\left( {a + b + c} \right) = 8\left( {a + b + c} \right) = 8.62 = 496\\ \Rightarrow \max P = 496\\ ' = ' \Leftrightarrow \left\{ \begin{array}{l} 5a + 3b = a + 7b\\ 5b + 3c = b + 7c\\ 5c + 3a = c + 7a \end{array} \right. \Leftrightarrow a = b = c\\ \Rightarrow a = b = c = \dfrac{{62}}{3} \end{array}$
