<s>[INST] Find minimum value of $P=\sum \frac{a}{a^2+b^3}$ If $a,b,c$ be non-negative numbers and $a+b+c=3$, then find minima value of function $$P=\frac{a}{a^2+b^3}+\frac{b}{b^2+c^3}+\frac{c}{c^2+a^3}$$

Here is my idea.
By WA i see that if the equality occurs  $a=b=c=1$, we will get $\text {Min}_P=1.5$. Indeed i will prove $P\ge 1.5$ by BW and C-S
We have: 
$$\text{L.H.S}=3\sum \frac{a^2}{a^3\sum a+3ab^3}\ge \frac{3\left(\sum a\right)^2}{\left(a+b+c\right)\left(\sum a^3\right)+3\sum ab^3}$$
Or $$2\left(\sum a\right)^4\ge 9\left(\sum a\right)\left(\sum a^3\right)+27\left(\sum ab^3\right)$$
I tried BW here but failed.The last inequality is wrong. Help me.
 [/INST] By the Andreas's hint it remains to prove that
$$\sum_{cyc}\frac{a}{a^2+b^3}\geq\frac{1}{3}.$$
Now, by C-S
$$\sum_{cyc}\frac{a}{a^2+b^3}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^3+b^3a)}.$$
Id est, it's enough to prove that
$$(a+b+c)^4\geq(a+b+c)(a^3+b^3+c^3)+3(a^3c+b^3a+c^3b)$$ or
$$\sum_{cyc}(3a^3b+6a^2b^2+12a^2bc)\geq0,$$ which is obvious.
Thus, $\frac{1}{3}$ is a minimal value.
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