อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Influenza_Mathematics
only cauchy and am-gm
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Well, from the fact
$\sum_{cyc}^{}(\frac{1}{a}-\frac{1}{b})^2 \geqslant 0$
$\Leftrightarrow \sum_{cyc}^{}\frac{1}{a^2}\geqslant \sum_{cyc}^{}\frac{1}{ab}$
$\Leftrightarrow \sum_{cyc}^{}\frac{3}{a^2} \geqslant \sum_{cyc}^{}\frac{1}{a^2}+2\sum_{cyc}^{}\frac{1}{ab}$
$\Leftrightarrow \sum_{cyc}^{}\frac{3}{a^2}\geqslant (\sum_{cyc}^{}\frac{1}{a})^2$
$\Leftrightarrow \frac{\sum_{cyc}^{}\frac{1}{a^2}}{3}\geqslant \frac{(\sum_{cyc}^{}\frac{1}{a})^2}{9}$
then do the same thing as #2