อ้างอิง:
ข้อความเดิมเขียนโดยคุณ BLACK-Dragon
Show that for all positive a,b,c
$$(\dfrac{a+2b}{a+2c})^3+(\dfrac{b+2c}{b+2a})^3+(\dfrac{c+2a}{c+2b})^3 \ge 3$$
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\[\sum_{cyc}(\frac{a+2b}{a+2c})^3\ge 3(\frac{1}{3}\sum_{cyc}\frac{a+2b}{a+2c})^3=3(\frac{1}{3}\sum_{cyc}\frac{a}{a+2c}+\frac{1}{3}\sum_{cyc}\frac{2b}{a+2c})^3\ge 3(\frac{1}{3}(\frac{(a+b+c)^2}{\sum_{cyc}a^2+2\sum_{cyc}ac})+\frac{1}{3}(\frac{2(a+b+c)^2}{\sum_{cyc}a^2+2\sum_{cyc}ac}))^3=3\]