<s>[INST] Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$ Let $a,\,b,\,c$ are non-negative such that $ab+bc+ca>0.$ Prove that
$$\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}. \quad (1)$$
Note. Because
$$\sum \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)} = 2.$$
So, from $(1)$ we get know inequality of Darij Grinberg
$$\frac{a(b+c)}{b^2+bc+c^2}+\frac{b(c+a)}{c^2+ca+a^2}+\frac{c(a+b)}{a^2+ab+b^2} \geqslant 2.$$
My proof use sum of squares method.
 [/INST] Thank @Michael Rozenberg :D We write inequality as
$$f(a) = 3(b+c)a^3-(5b^2+2bc+5c^2)a^2+(b+c)(b^2+bc+c^2)a+(b^2+bc+c^2)(b-c)^2 \geqslant 0.$$
But
$$\left(a+\frac{7b+7c}{4}\right) \cdot f(a)=\frac{3}{4}(b+c)(2a^2-b^2-c^2-ab+2bc-ca)^2+\frac{5bc(b+c)(b-c)^2}{4}$$
$$+\frac{17abc[(a-b)^2+(a-c)^2]}{4}+\frac{25a(b^2+c^2-ab-ca)^2}{8}+\left(\frac{a}{8}+b+c\right)(b-c)^2(b+c-a)^2.$$
So $f(a) \geqslant 0.$
</s>