<s>[INST] Prove inequality: $\sum \frac{a^4}{a^3+b^3} \ge \frac{a+b+c}{2}$ 
Prove inequality with $a,b,c >0$ $$\frac{a^4}{a^3+b^3} +
 \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}$$

I tried the inequality: $\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c$, but seem like it doesn't help
 [/INST] There is an extensive discussion of this inequality here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=23113
I like  the sum of squares method the best:
\begin{align}
& \sum_{cyc}{\frac{a^4}{a^3+b^3}}-\frac{1}{2}\sum_{cyc}{a} \\
& =\frac{(a-b)^2(b-c)^2(c-a)^2(ab+ac+bc)^2}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)} \\
& +\sum_{cyc}{\frac{c^2(b-c)^2(2c^3(a^3+b^3)+c^2a^2(a+b)^2+2ca^4b+a^4(a^2+b^2))}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)}} \\
& \geq 0
\end{align}
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