<s>[INST] 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!
 [/INST] 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!
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