C/m hằng đẳng thức: a. a^3 + b^3 >= a^2.b + a.b^2 (a,b >= 0) b. a^5 + b^5 >= a^3.b^2 + a^2.b^3 (a,b >= 0) 30/07/2021 Bởi Skylar C/m hằng đẳng thức: a. a^3 + b^3 >= a^2.b + a.b^2 (a,b >= 0) b. a^5 + b^5 >= a^3.b^2 + a^2.b^3 (a,b >= 0)
Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\\left( {{a^3} + {b^3}} \right) – \left( {{a^2}b + {b^2}a} \right)\\ = \left( {{a^3} – {a^2}b} \right) + \left( {{b^3} – {b^2}a} \right)\\ = {a^2}\left( {a – b} \right) + {b^2}\left( {b – a} \right)\\ = \left( {a – b} \right)\left( {{a^2} – {b^2}} \right)\\ = \left( {a – b} \right)\left( {a – b} \right)\left( {a + b} \right)\\ = {\left( {a – b} \right)^2}\left( {a + b} \right) \ge 0,\forall a,b \ge 0\\ \Rightarrow {a^3} + {b^3} \ge {a^2}b + a{b^2}\\b,\\\left( {{a^5} + {b^5}} \right) – \left( {{a^3}{b^2} + {a^2}{b^3}} \right)\\ = \left( {{a^5} – {a^3}{b^2}} \right) + \left( {{b^5} – {a^2}{b^3}} \right)\\ = {a^3}\left( {{a^2} – {b^2}} \right) + {b^3}\left( {{b^2} – {a^2}} \right)\\ = \left( {{a^2} – {b^2}} \right)\left( {{a^3} – {b^3}} \right)\\ = \left( {a – b} \right)\left( {a + b} \right)\left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)\\ = {\left( {a – b} \right)^2}\left( {a + b} \right)\left( {{a^2} + ab + {b^2}} \right) \ge 0,\forall a,b \ge 0\\ \Rightarrow {a^5} + {b^5} \ge {a^3}{b^2} + {a^2}{b^3}\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {{a^3} + {b^3}} \right) – \left( {{a^2}b + {b^2}a} \right)\\
= \left( {{a^3} – {a^2}b} \right) + \left( {{b^3} – {b^2}a} \right)\\
= {a^2}\left( {a – b} \right) + {b^2}\left( {b – a} \right)\\
= \left( {a – b} \right)\left( {{a^2} – {b^2}} \right)\\
= \left( {a – b} \right)\left( {a – b} \right)\left( {a + b} \right)\\
= {\left( {a – b} \right)^2}\left( {a + b} \right) \ge 0,\forall a,b \ge 0\\
\Rightarrow {a^3} + {b^3} \ge {a^2}b + a{b^2}\\
b,\\
\left( {{a^5} + {b^5}} \right) – \left( {{a^3}{b^2} + {a^2}{b^3}} \right)\\
= \left( {{a^5} – {a^3}{b^2}} \right) + \left( {{b^5} – {a^2}{b^3}} \right)\\
= {a^3}\left( {{a^2} – {b^2}} \right) + {b^3}\left( {{b^2} – {a^2}} \right)\\
= \left( {{a^2} – {b^2}} \right)\left( {{a^3} – {b^3}} \right)\\
= \left( {a – b} \right)\left( {a + b} \right)\left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= {\left( {a – b} \right)^2}\left( {a + b} \right)\left( {{a^2} + ab + {b^2}} \right) \ge 0,\forall a,b \ge 0\\
\Rightarrow {a^5} + {b^5} \ge {a^3}{b^2} + {a^2}{b^3}
\end{array}\)