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ข้อความเดิมเขียนโดยคุณ จูกัดเหลียง
Let $a,b,c>0$ and $a+b+c=3$ Prove that $$\frac{1}{\Big((a-b)^2+3bc+3ca\Big)^2}+\frac{1}{\Big((b-c)^2+3ca+3ab\Big)^2}+\frac{1}{\Big((c-a)^2+3ab+3bc\Big)^2}\ge \frac{1}{12}$$
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$\displaystyle \frac{1}{\Big((a-b)^2+3bc+3ca\Big)^2}+\frac{1}{\Big((b-c)^2+3ca+3ab\Big)^2}+\frac{1}{\Big((c-a)^2+3ab+3bc\Big)^2}$
$\displaystyle \ge \frac{1}{3} \Big( \frac{1}{(a-b)^2+3bc+3ca}+\frac{1}{(b-c)^2+3ca+3ab}+\frac{1}{(c-a)^2+3ab+3bc} \Big)^2$
$\displaystyle \ge \frac{1}{3} \Big( \frac{9}{2(a^2+b^2+c^2)+4(ab+bc+ca)}\Big)^2 = \frac{1}{3} \cdot \Big( \frac{9}{2\cdot 3^2} \Big)^2 = \frac{1}{12}$