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