Lời giải:Ta có:
\(\begin{array}{l}
{A^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \left( { - \frac{\pi }{2} - \frac{\pi }{3}} \right)\\
\Leftrightarrow {5^5} = A_1^2 + A_2^2 - \sqrt 3 {A_1}{A_2} \Leftrightarrow {5^2} = {\left( {{A_1} + {A_2}} \right)^2} - 3,73{A_1}{A_2}\\
\Rightarrow {A_1}{A_2} = \frac{{{{\left( {{A_1} + {A_2}} \right)}^2} - {5^2}}}{{3,732}}
\end{array}\)
Theo bất đẳng thức cô si:
\(\begin{array}{l}
{A_1} + {A_2} \ge 2\sqrt {{A_1}{A_2}} \Rightarrow {A_1}{A_2} \le \frac{{{{\left( {{A_1} + {A_2}} \right)}^2}}}{4}\\
\Rightarrow \frac{{{{\left( {{A_1} + {A_2}} \right)}^2} - {5^2}}}{{3,732}} \le \frac{{{{\left( {{A_1} + {A_2}} \right)}^2}}}{4} \Rightarrow {\left( {{A_1} + {A_2}} \right)^2} - 3,732\frac{{{{\left( {{A_1} + {A_2}} \right)}^2}}}{4} \le {5^2}\\
\Rightarrow 0,067{\left( {{A_1} + {A_2}} \right)^2} \le {5^2} \Rightarrow \left( {{A_1} + {A_2}} \right) \le 19,3
\end{array}\)
