[จูกัดเหลียง]

ข้อความเดิมเขียนโดยคุณ nooonuii

ต่อให้อีก

21. $a,b,c>0$

$$\dfrac{a^2+(b-c)^2}{a^2+(b+c)^2}+\dfrac{b^2+(c-a)^2}{b^2+(c+a)^2}+\dfrac{c^2+(a-b)^2}{c^2+(a+b)^2}\geq\dfrac{3}{5}$$

$$\Leftrightarrow 2(\sum_{cyc} a^2)(\sum_{cyc} \frac{1}{a^2+(b+c)^2}) \ge \frac{18}{5}$$
and by AM.-HM. $$\sum_{cyc} \frac{1}{a^2+(b+c)^2} \ge \frac{9}{\sum_{cyc} a^2+ \sum_{cyc} (a+b)^2}=\frac{9}{3(\sum_{cyc} a^2)+2\sum_{cyc} ab} \ge \frac{9}{5\sum_{cyc} a^2}$$