Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$

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.

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. $\qquad \Box$