Diễn Đàn MathScope - Xem bài viết đơn

supermouse · 10-03-2011, 10:22 PM

Bài 2: VT
$\begin{array}{l}
\Leftrightarrow a - \frac{{2a{b^2}}}{{a + 2{b^2}}} + b - \frac{{2b{c^2}}}{{b + 2{c^2}}} + c - \frac{{2c{a^2}}}{{c + 2{a^2}}} \ge 3 - \sum {\frac{{2a{b^2}}}{{3\sqrt[3]{{a{b^4}}}}}} = 3 - \frac{2}{3}\sum {\sqrt[3]{{{a^2}{b^2}}}} \\
ab + ab + 1 \ge 3\sqrt[3]{{{a^2}{b^2}}};\frac{{{a^2} + {b^2} + 1}}{2} \ge \frac{3}{2}\sqrt[2]{{{a^2}{b^2}}} \\
\Rightarrow {(a + b + c)^2} + \frac{9}{2} \ge \frac{9}{2}\sum {\sqrt[3]{{{a^2}{b^2}}}} \\
\Rightarrow \frac{3}{2}\sum {\sqrt[3]{{{a^2}{b^2}}}} \le 2 \\
\end{array} $
Ta có đfcm

10-03-2011, 10:22 PM	#2
supermouse +Thành Viên+ Tham gia ngày: Mar 2010 Đến từ: diamond planet Bài gởi: 85 Thanks: 10 Thanked 45 Times in 29 Posts	Bài 2: VT $\begin{array}{l} \Leftrightarrow a - \frac{{2a{b^2}}}{{a + 2{b^2}}} + b - \frac{{2b{c^2}}}{{b + 2{c^2}}} + c - \frac{{2c{a^2}}}{{c + 2{a^2}}} \ge 3 - \sum {\frac{{2a{b^2}}}{{3\sqrt[3]{{a{b^4}}}}}} = 3 - \frac{2}{3}\sum {\sqrt[3]{{{a^2}{b^2}}}} \\ ab + ab + 1 \ge 3\sqrt[3]{{{a^2}{b^2}}};\frac{{{a^2} + {b^2} + 1}}{2} \ge \frac{3}{2}\sqrt[2]{{{a^2}{b^2}}} \\ \Rightarrow {(a + b + c)^2} + \frac{9}{2} \ge \frac{9}{2}\sum {\sqrt[3]{{{a^2}{b^2}}}} \\ \Rightarrow \frac{3}{2}\sum {\sqrt[3]{{{a^2}{b^2}}}} \le 2 \\ \end{array} $ Ta có đfcm __________________ *NEVER GIVE UP*☺☺☺☺☺☺☺☺