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!

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.
