อ้างอิง:
ข้อความเดิมเขียนโดยคุณ จูกัดเหลียง
44.Let $a,b,c\in \mathbb{R^+}$
Prove that $$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\ge 3$$
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$a+b+c\le a^2+b^2+c^2$
$$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\ge \sum_{cyc} \frac{a^2+b^2}{a^2+b^2+c^2-c}$$
It's Remain to show that $$ \sum_{cyc} \frac{a^2+b^2}{a^2+b^2+c^2-c}\ge 3\leftrightarrow \sum_{cyc} \frac{a-a^2}{3-a}\ge 0$$
Use $f(x)=\frac{x-x^2}{3-x}$ Then $f$ is Convex by $f^{"}(x)=\frac{-6 (2x^2-7x-5)}{(3-x)^3}>0\leftrightarrow x<\frac{5}{2}$
Then It's cleared!!!