Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$

Question

Give $a,b,c>0$ and $ab^2+bc^2+ca^2=1$

Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$

Answer 1

$$P = \sum_{cyc}\left(a^3-2\frac{a^3b}{a+2b} \right) \ge \sum_{cyc}\left(a^3-2\frac{a^3b}{3\sqrt[3]{ab^2}} \right)=\sum_{cyc}\left(a^3-\frac23 a^{8/3}b^{1/3}\right)$$

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.

Answer 2

For $a=b=c=\frac{1}{\sqrt[3]3}$ we get a value $\frac{1}{3}$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$\sum_{cyc}\frac{a^4}{a+2b}\geq\frac{ab^2+bc^2+ca^2}{3}$$ or $$\sum_{cyc}\left(\frac{a^4}{a+2b}-\frac{ab^2}{3}\right)\geq0$$ or $$\sum_{cyc}\frac{a(a-b)(3a^2+3ab+2b^2)}{a+2b}\geq0$$ or $$\sum_{cyc}\left(\frac{a(a-b)(3a^2+3ab+2b^2)}{a+2b}-\frac{8}{9}(a^3-b^3)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(19a^2+22ab+16b^2)}{a+2b}\geq0.$$ Done!