prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
Let $p=a+b+c=1, \; q=ab+bc+ca, \; r=abc.$ We need to prove $$5(ab+bc+ca-abc) \geqslant 4(1+ab+bc+ca)(ab+bc+ca-3abc),$$ equivalent to $$5(q-r) \geqslant 4(1+q)(q-3r),$$ or $$(12q+7)r \geqslant q(4q-1).$$ If $4q-1 < 0,$ then $$(12q+7)r > 0 > q(4q-1).$$ If $4q-1 \geqslant 0,$ from Schur inequality $$(a+b+c)^3+9abc \geqslant 4(a+b+c)(ab+bc+ca),$$ we get $$r \geqslant \frac{p(4q-p^2)}{9} = \frac{4q-1}{9}.$$ It's remain to prove that $$(12q+7) \cdot \frac{4q-1}{9} \geqslant q(4q-1),$$ or $$\frac{(3q+7)(4q-1)}{9} \geqslant 0.$$ Which is true. The proof is completed.
Proof by full expanding.
After your work we need to prove that: $$5\left(\left(\sum_{cyc}a\right)^3\sum_{cyc}ab-abc\left(\sum_{cyc}a\right)^2\right)\geq4\left(\left(\sum_{cyc}a\right)^2+\sum_{cyc}ab\right)\left(\sum_{cyc}a\sum_{cyc}ab-3abc\right)$$ or $$5\sum_{cyc}(a^2+2ab)\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)\geq4\sum_{cyc}(a^2+3ab)\sum_{cyc}(a^2b+a^2c)$$ or $$\sum_{cyc}(a^2+2ab)\sum_{cyc}(a^2b+a^2c)+10abc\sum_{cyc}(a^2+2ab)\geq4\sum_{cyc}ab\sum_{cyc}(a^2b+a^2c)$$ or $$\sum_{cyc}(a^2-2ab)\sum_{cyc}(a^2b+a^2c)+10abc\sum_{cyc}(a^2+2ab)\geq0$$ or $$\sum_{cyc}(a^4b+a^4c+a^3b^2+a^3c^2+2a^2b^2c-2a^3b^2-2a^3c^2-4a^3bc-4a^2b^2c+$$ $$+10a^3bc+10a^2b^2c)\geq0$$ or $$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2+6a^3bc+8a^2b^2c)\geq0,$$ which is true by Muirhed.
Of course, I could write a last line only, but I showed, how we can get it: three minutes of work!
Also, $uvw$ helps here very well.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, after homogenization your last inequality it's $f(w^3)\geq0,$ where $f$ is a linear function,
which says that it's enough to prove it for the extreme value of $w^3$,
which by $uvw$ happens in the following cases.
- $w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
Thus, we need to prove that:$$5ab\geq4(1+ab)ab$$ or $$ab\leq\frac{1}{4},$$ which is true by AM-GM: $$ab\leq\left(\frac{a+b}{2}\right)=\frac{1}{4}.$$
- Two variables are equal.
Let $b=a$ and $c=1-2a$, where $0<a<\frac{1}{2}.$
After this substitution we need to prove that: $$a(1-2a)(1-a)(18a^2-3a+1)\geq0,$$ which is obvious.