prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$ 
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$

Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is homogenous we take $a+b+c=1$.
or we have to prove (i am skipping the steps as it is just algebra) :
$$ 5(ab+bc+ca-abc)\ge 4(1+ab+bc+ca)(ab+bc+ca-3abc)$$
But i am not able to prove this by expanding.
How do i proceed?
Other methods are welcome!

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.
