Đáp án:
$\begin{array}{l}
Bunhia:{\left( {a.x + b.y} \right)^2} \le \left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right)\\
\Rightarrow \left( {a.x + b.y} \right) \le \sqrt {\left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right)} \\
\Rightarrow \sqrt {a + b} + \sqrt {b + c} + \sqrt {c + a} \\
= 1.\sqrt {a + b} + 1.\sqrt {b + c} + 1.\sqrt {c + a} \le \sqrt {\left( {1 + 1 + 1} \right).\left( {a + b + b + c + c + a} \right)} \\
\Rightarrow \sqrt {a + b} + \sqrt {b + c} + \sqrt {c + a} \le \sqrt {3.2\left( {a + b + c} \right)} = \sqrt 6 \\
Dấu\, = \,xảy\,ra \Leftrightarrow a = b = c = \frac{1}{3}
\end{array}$
