Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {a - 1} \right)^2} \ge 0 \Leftrightarrow {a^2} - 2a + 1 \ge 0 \Leftrightarrow {a^2} + 1 \ge 2a\\
{\left( {b - 1} \right)^2} \ge 0 \Leftrightarrow {b^2} - 2b + 1 \ge 0 \Leftrightarrow {b^2} + 1 \ge 2b\\
{\left( {c - 1} \right)^2} \ge 0 \Leftrightarrow {c^2} - 2c + 1 \ge 0 \Leftrightarrow {c^2} + 1 \ge 2c\\
\frac{a}{{1 + {b^2}}} = \frac{{a\left( {1 + {b^2}} \right) - a{b^2}}}{{1 + {b^2}}} = a - \frac{{a{b^2}}}{{1 + {b^2}}}\\
ab + bc + ca \le {a^2} + {b^2} + {c^2} \Rightarrow ab + bc + ca \le \frac{{{{\left( {a + b + c} \right)}^2}}}{3} = 3\\
A = \frac{a}{{1 + {b^2}}} + \frac{b}{{1 + {c^2}}} + \frac{c}{{1 + {a^2}}}\\
= \left( {a - \frac{{a{b^2}}}{{1 + {b^2}}}} \right) + \left( {b - \frac{{b{c^2}}}{{1 + {c^2}}}} \right) + \left( {c - \frac{{c{a^2}}}{{1 + {a^2}}}} \right)\\
\ge \left( {a + b + c} \right) - \left( {\frac{{a{b^2}}}{{2b}} + \frac{{b{c^2}}}{{2c}} + \frac{{c{a^2}}}{{2a}}} \right)\\
= 3 - \frac{1}{2}\left( {ab + bc + ca} \right) \ge 3 - \frac{1}{2}.3 = \frac{3}{2}
\end{array}\)
Dấu '=' xảy ra khi và chỉ khi a=b=c
