[Devilbat]

ทำข้อนี้ตอนแรกไม่คิดว่าจะใช้ Cauchy reverse โจทย์สวยดีครับ
$\frac{a^2}{a+2b^3}+\frac{b^2}{b+2c^3}+\frac{c^2}{c+2a^3}\ge1$ is equivalent to
$\sum_{cyc}a-\sum_{cyc}\frac{2ab^3}{a+2b^3}\ge1$ or
$\sum_{cyc}\frac{ab^3}{a+2b^3}\le 1$ By A.M-G.M , we only need to prove that
$a^{2/3}b+b^{2/3}c+c^{2/3}a\le3$ By A.M-G.M, we obtain
$\sum_{cyc}3a^{2/3}b\le\sum_{cyc}b+ab+ab=\sum_{cyc}a+2\sum_{cyc}ab\le3(a+b+c)=9$
and the desired result follows.