Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$ Give $a,b,c>0$ and $ab^2+bc^2+ca^2=1$
Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$

$$P =  \sum_{cyc}\left(a^3-2\frac{a^3b}{a+2b} \right) \ge \sum_{cyc}\left(a^3-2\frac{a^3b}{3\sqrt[3]{ab^2}} \right)=\sum_{cyc}\left(a^3-\frac23 a^{8/3}b^{1/3}\right)$$
Now from Rearrangement (or using AM-GM), we have $\sum_{cyc} a^3 \ge  \sum_{cyc} a^{8/3}b^{1/3}$ and $\sum_{cyc} a^3 \ge \sum_{cyc} ab^2$, and using these we have
$$P \ge \frac13 \sum_{cyc} ab^2=\frac13$$  
as equality is reached when $a=b=c=\frac1{\sqrt3}$, this is indeed the minimum.
