how to solve $\displaystyle \frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$ 
$a,b,c>0,a+b+c=3,$
prove that:
$$\frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$$


With $\sum $ denoting cyclic sum, using Cauchy Schwarz inequality:
$$\sum \frac{a^4}{a(a^2+2b+c)} \ge \frac{(a^2+b^2+c^2)^2}{\sum a^3+3\sum ab}$$
So it is enough to show 
$$4(a^2+b^2+c^2)^2 \ge (a+b+c)\sum a^3+(a+b+c)^2\sum ab$$
$$\iff 3\sum a^4 +6\sum a^2b^2 \ge 2\sum ab(a^2+b^2)+5abc\sum a$$
which follows from Schur $\sum a^4 + abc\sum a\ge \sum ab(a^2+b^2)$ and the AM-GMs $\sum a^2b^2 \ge \sum a^2bc$ and $2\sum a^4 \ge \sum ab(a^2+b^2)$. 
