Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$

Question

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.

Answer 1

BW helps!

We need to prove that: $$3(b+c)a^3-(5b^2+2bc+5c^2)a^2+(b+c)(b^2+bc+c^2)a+(b-c)^2(b^2+bc+c^2)\geq0.$$

Thus, since our inequality is symmetric respect to $b$ and $c$, we have two cases only:

1. $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Thus, $u\geq0$, $v\geq0$ and we need to prove that: $$3(u^2+v^2)a^2+(u+v)(4u^2-5uv+4v^2)a+(u-v)^2(u^2+uv+v^2)\geq0,$$ which is obvious;

2. $b=\min\{a,b,c\}$, $a=b+u$ and $c=b+v$.

Thus, we need to prove that: $$3(2u^2-2uv+v^2)b^2+(6u^3-3u^2v-5uv^2+4v^2)b+v(3u+v)(u-v)^2\geq0,$$ which is obvious again.

Answer 2

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 $f(a) \geqslant 0.$

Answer 3

We have$:$ $$48\, \left( b+c \right) ^{2} \left( {b}^{2}+bc+{c}^{2} \right) \left( {a}^{2}+ab+ac+{b}^{2}+bc+{c}^{2} \right) \cdot (\text{LHS}-\text{RHS})$$ $$=48\,a ( b+c ) \Big[b^2+c^2-a(b+c)\Big] ^{2}+4\,(b-c)^2 \Big[ 2\, a(b+c)-(2b^2+bc+2c^2) \Big] ^{2}$$

$$+4\,bc \left( 8\,{b}^{2}+7\,bc+8\,{c}^{2} \right) \left( b-c \right) ^ {2} \geqq 0.$$