Trích:
Nguyên văn bởi supermouse  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 |
Pãi chăng đây là cauchy schwarz
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]