Đáp án:
\[{A_{\min }} = \frac{1}{2} \Leftrightarrow a = b = \frac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = a\left( {{a^2} + 2b} \right) + b\left( {{b^2} - a} \right)\\
= {a^3} + 2ab + {b^3} - ab\\
= \left( {{a^3} + {b^3}} \right) + ab\\
= \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) + ab\\
= 1.\left( {{a^2} - ab + {b^2}} \right) + ab\\
= {a^2} + {b^2}\\
{\left( {a - b} \right)^2} \ge 0,\,\,\,\forall a,b\\
\Rightarrow {a^2} + {b^2} \ge 2ab\\
\Leftrightarrow 2\left( {{a^2} + {b^2}} \right) \ge {\left( {a + b} \right)^2}\\
\Leftrightarrow {a^2} + {b^2} \ge \frac{1}{2}\\
\Leftrightarrow A \ge \frac{1}{2}\\
\Rightarrow {A_{\min }} = \frac{1}{2} \Leftrightarrow a = b = \frac{1}{2}
\end{array}\)
