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!

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.
