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:
$$\frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$$

 A: Starting from the Cauchy-Schwarz inequality:
$$(x_1^2+y_1^2+z_1^2)( x_2^2+y_2^2+z_2^2)\geqslant (x_1x_2+y_1y_2+z_1z_2)^2;x_i,y_i\in R$$  
Setting  
$$x_1^2=a^2+2b+c, y_1^2=b^2+2c+a, z_1^2=c^2+2a+b $$
$$x_2^2=\frac{a^3}{a^2+2b+c}, y_2^2=\frac{b^3}{b^2+2c+a}, z_2^2=\frac{c^3}{c^2+2a+b} $$  
Which gives
$$(a^2+b^2+c^2+3(a+b+c))( \frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}))\geqslant (a^{\frac{3}{2}}+ b^{\frac{3}{2}}+ c^{\frac{3}{2}})^2$$  
Therefore, if the following inequality holds, the original problem is solved.
$$\frac{(a^{\frac{3}{2}}+ b^{\frac{3}{2}}+ c^{\frac{3}{2}})^2}{ a^2+b^2+c^2+3(a+b+c)}\geqslant\frac{3}{4} ;\quad s.t\quad a+b+c=3$$ 
Since $a,b,c>0$ we can use the following change of variables;
$$u=a^\frac{1}{2}, v=b^\frac{1}{2}, w=c^\frac{1}{2}$$  
$$\Rightarrow\frac{(u^3+v^3+w^3)^2}{u^4+v^4+w^4+3(u^2+v^2+w^2)}\geqslant\frac{3}{4}$$
$$\quad s.t\quad u^2+v^2+w^2=3$$  
In order to prove the above inequality, it suffices to show that the optimum value of the following optimization problem is zero;
$$minf(u,v,w)=4(u^3+v^3+w^3)^2-3(u^4+v^4+w^4+3(u^2+v^2+w^2))$$
$$\quad s.t\quad u,v,w>0; u^2+v^2+w^2=3$$  
Using the Lagrange multiplier method, the solution is  
$$u=v=w=1,f_{min}=0$$  
Therefore
$$f(u,v,w)\geqslant0\Rightarrow\frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geqslant\frac{(a^{\frac{3}{2}}+ b^{\frac{3}{2}}+ c^{\frac{3}{2}})^2}{ a^2+b^2+c^2+3(a+b+c)}\geqslant\frac{3}{4}$$
Q.E.D.
A: 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 $2\sum a^4 \ge \sum ab(a^2+b^2)$. 
