Đáp án:
Áp dụng bđt Bunhia ta có:
$\begin{array}{l}
{\left( {3x + 4y} \right)^2} \le \left( {{3^2} + {4^2}} \right)\left( {{x^2} + {y^2}} \right)\\
\Rightarrow {\left( {3x + 4y} \right)^2} \le 25.1 = 25\left( {do:{x^2} + {y^2} = 1} \right)\\
\Rightarrow - 5 \le 3x + 4y \le 5\\
Dấu = xảy\,ra \Leftrightarrow \left\{ \begin{array}{l}
\frac{x}{3} = \frac{y}{4}\\
{x^2} + {y^2} = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{3}{5}\\
y = \frac{4}{5}
\end{array} \right.
\end{array}$
