Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\left( {{a^5} + {b^5}} \right) - \left( {{a^2} + {b^2}} \right)\left( {{a^3} + {b^3}} \right)\\
= 2\left( {{a^5} + {b^5}} \right) - \left( {{a^5} + {a^2}{b^3} + {a^3}{b^2} + {b^5}} \right)\\
= {a^5} + {b^5} - {a^2}{b^3} - {a^3}{b^2}\\
= {a^3}\left( {{a^2} - {b^2}} \right) + {b^3}\left( {{b^2} - {a^2}} \right)\\
= \left( {{a^2} - {b^2}} \right)\left( {{a^3} - {b^3}} \right)\\
= \left( {a - b} \right)\left( {a + b} \right)\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= {\left( {a - b} \right)^2}\left( {a + b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= {\left( {a - b} \right)^2}\left( {a + b} \right)\left[ {{{\left( {a + \frac{1}{2}b} \right)}^2} + \frac{3}{4}{b^2}} \right] \ge 0,\,\,\,\,\,\forall a + b \ge 0\\
\Rightarrow 2\left( {{a^5} + {b^5}} \right) \ge \left( {{a^2} + {b^2}} \right)\left( {{a^3} + {b^3}} \right),\forall a + b \ge 0
\end{array}\)
