Chứng minh rằng nếu a ≥ 0 và b > 0 thì
\(\frac{{a + b}}{2}.\frac{{{a^2} + {b^2}}}{2} \le \frac{{{a^3} + {b^3}}}{2}\)
Ta có:
\(\begin{array}{l}
\frac{{a + b}}{2}.\frac{{{a^2} + {b^2}}}{2} \le \frac{{{a^3} + {b^3}}}{2}\\
\Leftrightarrow {a^3} + a{b^2} + {a^2}b + {b^3} \le 2{a^3} + 2{b^3}\\
\Leftrightarrow {a^3} - a{b^2} - {a^2}b + {b^3} \ge 0\\
\Leftrightarrow \left( {a - b} \right)\left( {{a^2} - {b^2}} \right) \ge 0\\
\Leftrightarrow {\left( {a - b} \right)^2}\left( {a + b} \right) \ge 0\left( {ld} \right)
\end{array}\)
Vậy \(\frac{{a + b}}{2}.\frac{{{a^2} + {b^2}}}{2} \le \frac{{{a^3} + {b^3}}}{2}\)
-- Mod Toán 10
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