Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$ 
Prove that (where $a,b,c>0$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$

I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.
Background:
Here however is a very similar problem

If $a,b,c>0$ then prove that $$\sum_{cyc} \frac{a(a^2+bc)}{b+c}\ge a^2+b^2+c^2$$

There is a beautiful proof of  this by Michael Rozenberg (arqady)
WLOG $a\ge b\ge c$ rewrite the inequality as $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\ge 0$$ but  $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\geq\frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{a+c}=$$
$$=(a-b)(\frac{a(a-c)}{b+c}-\frac{b(b-c)}{a+c})\geq0$$
However this method doesnt work...
I am looking for a clean and smooth proof (without using BW,uvw or complete expanding)
I am not planning to disclose how I proved the inequality as it may spoil the fun!
 A: Suppose $c = \min\{a,b,c\},$ then
$$\sum \frac{a^3+abc}{b+c}-(a^2+b^2+c^2) = \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a}+\frac{c(c-a)(c-b)}{a+b}$$
$$\geqslant \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a} = \frac{(a-b)^2(a^2+ab+b^2-c^2)}{(b+c)(c+a)} \geqslant 0.$$
Therefore
$$\sum \frac{a^3+abc}{b+c} \geqslant a^2+b^2+c^2. \qquad (1)$$
Now, we write the inequality as
$$\sum \frac{a^3+abc}{b+c} + abc \sum \frac{1}{b+c} \geqslant \frac{(a+b+c)^2}{2}.$$
By the Cauchy-Schwarz inequality, we have
$$\sum \frac{1}{b+c} \geqslant \frac{9}{2(a+b+c)}. \qquad (2)$$
From $(1)$ and $(2),$ we need to prove
$$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)} \geqslant \frac{(a+b+c)^2}{2},$$
equivalent to
$$a^2+b^2+c^2+\frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
This is Schur inequality. The proof is completed.
A: We want to show
$$
\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}
$$
The following steps mainly show how this inequality can be maximally simplified.
The final steps then have been proved by the OP Albus Dumbledore himself, and others.
Due to homogeneity, let $a+b+c =1$, then equivalently
$$
\sum_{cyc}  \frac{(a-1+1)(a^2+2bc)}{1-a} \ge \frac{1}{2} $$
$$-\sum_{cyc} (a^2+2bc) + \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$
$$-(a+b+c)^2+ \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$
$$\sum_{cyc} \frac{a^2-1+1+2bc}{1-a} \ge \frac{3}{2}
$$
$$-\sum_{cyc} (1+a)+\sum_{cyc} \frac{1+2bc}{1-a} \ge \frac{3}{2}
$$
$$\sum_{cyc} \frac{1+2bc}{b+c} \ge \frac{11}{2}
$$
Now we have
$$\frac{2bc}{1-a} =\frac{2bc}{b+c} = b+c -a-1  + \frac{1 -(a^2+b^2+c^2)}{b+c} $$
which leads to
$$(2 -(a^2+b^2+c^2))\sum_{cyc} \frac{1}{b+c} \ge \frac{15}{2}
$$
Now we have isolated a single sum, which is the main benefit of this answer.
This sum can be evaluated:
$$\sum_{cyc} \frac{1}{1-a} = \frac{3 -  (a^2+b^2+c^2)}{1 - 2abc - (a^2+b^2+c^2)}
$$
which leaves to show
$$(2 -(a^2+b^2+c^2))(3 -(a^2+b^2+c^2)) \ge \frac{15}{2} (1 - 2abc - (a^2+b^2+c^2))
$$
Let $x = a^2+b^2+c^2$ then we have to show
$$2x^2 + 5x + 30 abc \ge 3
$$
Note that by now, no single change has been made to the original question, since all transformations are equivalences.
The last inequality  can be proved (amongst other methods)  by Schur,  which has been done by the OP Albus Dumbledore himself, and others, see here. $\qquad \Box$
