Ta có: \[\left\{ \begin{array}{l}
{a^2} + b + c \ge a + b + c\\
{b^2} + c + a \ge b + c + a\\
{c^2} + a + b \ge c + a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\frac{1}{{{a^2} + b + c}} \le \frac{1}{{a + b + c}}\\
\frac{1}{{{b^2} + c + a}} \le \frac{1}{{b + c + a}}\\
\frac{1}{{{c^2} + a + b}} \le \frac{1}{{c + a + b}}
\end{array} \right.\]
\[\begin{array}{l}
\Rightarrow \frac{1}{{{a^2} + b + c}} + \frac{1}{{{b^2} + c + a}} + \frac{1}{{{c^2} + a + b}} \le \frac{1}{{a + b + c}} + \frac{1}{{b + c + a}} + \frac{1}{{c + a + b}} = 1\\
(a + b + c = 3)
\end{array}\]
Dấu "=" xảy ra \[ \Leftrightarrow \left\{ \begin{array}{l}
{a^2} = a;{b^2} = b;{c^2} = c\\
a + b + c = 3
\end{array} \right.\]
$ \Leftrightarrow a = b = c = \frac{1}{3}$
@congchua: Bài này sai nhé bạn