Đáp án:

\[{A_{\min }} = \dfrac{{\sqrt 3 }}{2} \Leftrightarrow a = b = c = \dfrac{{\sqrt 3 }}{3}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
{\left( {a - b} \right)^2} \ge 0,\,\,\forall a,b\\
{\left( {b - c} \right)^2} \ge 0,\,\,\,\forall b,c\\
{\left( {c - a} \right)^2} \ge 0,\,\,\,\,\forall c,a\\
 \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0,\,\,\,\forall a,b,c\\
 \Leftrightarrow {a^2} - 2ab + {b^2} + {b^2} - 2bc + {c^2} + {c^2} - 2ca + {a^2} \ge 0,\,\,\,\forall a,b,c\\
 \Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) \ge 2\left( {ab + bc + ca} \right),\,\,\,\,\,\forall a,b,c\\
 \Leftrightarrow {a^2} + {b^2} + {c^2} \ge ab + bc + ca\\
{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \ge 3.\left( {ab + bc + ca} \right) = 3\\
 \Rightarrow a + b + c \ge \sqrt 3 
\end{array}\)

Áp dụng bất đẳng thức Cô - si ta có:

\(\begin{array}{l}
\dfrac{{{a^2}}}{{b + c}} + \dfrac{{b + c}}{4} \ge 2.\sqrt {\dfrac{{{a^2}}}{{b + c}}.\dfrac{{b + c}}{4}}  = 2.\dfrac{a}{2} = a\\
\dfrac{{{b^2}}}{{c + a}} + \dfrac{{c + a}}{4} \ge 2.\sqrt {\dfrac{{{b^2}}}{{c + a}}.\dfrac{{c + a}}{4}}  = 2.\dfrac{b}{2} = b\\
\dfrac{{{c^2}}}{{a + b}} + \dfrac{{a + b}}{4} \ge 2.\sqrt {\dfrac{{{c^2}}}{{a + b}}.\dfrac{{a + b}}{4}}  = 2.\dfrac{c}{2} = c\\
 \Rightarrow \left( {\dfrac{{{a^2}}}{{b + c}} + \dfrac{{b + c}}{4}} \right) + \left( {\dfrac{{{b^2}}}{{c + a}} + \dfrac{{c + a}}{4}} \right) + \left( {\dfrac{{{c^2}}}{{a + b}} + \dfrac{{a + b}}{4}} \right) \ge a + b + c\\
 \Leftrightarrow \left( {\dfrac{{{a^2}}}{{b + c}} + \dfrac{{{b^2}}}{{c + a}} + \dfrac{{{c^2}}}{{a + b}}} \right) + \left( {\dfrac{{a + b + c}}{2}} \right) \ge a + b + c\\
 \Leftrightarrow A \ge \dfrac{{a + b + c}}{2} \ge \dfrac{{\sqrt 3 }}{2}
\end{array}\)

Dấu '=' xảy ra khi và chỉ khi: \(a = b = c = \dfrac{{\sqrt 3 }}{3}\)

Vậy \({A_{\min }} = \dfrac{{\sqrt 3 }}{2} \Leftrightarrow a = b = c = \dfrac{{\sqrt 3 }}{3}\)