For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$
where $k={\frac {27}{8}}+\frac{9\sqrt{3}}{4}$ is a Root Of $64{k}^{2}-432k-243=0.$
By computer (Maple) I found this inequality is equivalent to$:$ $$\sum \Big[ a+b+ ( 1-\frac{\sqrt {3}}{2} ) c\Big] \Big[ 2(a+b)- ( 1+\sqrt {3} ) c \Big] ^{2} ( \,a-b \,) ^{2} \geqslant 0$$
But I hope for alternative proof$?$
Best Answer
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality it's a linear inequality of $w^3$, which says that it's enough to prove our inequality for an extreme value of $w^3$, which by $uvw$ happens in the following cases.
Let $c=0$, $b=1$ and $a^2+1=2ua$.
Thus, $u\geq1$ and we need to prove here that $$\frac{a^2-a+1}{a}+\frac{ka}{(a+1)^2}\geq\frac{3}{8}+\frac{k}{3}$$ or $$2u-1+\frac{k}{2u+2}\geq\frac{3}{8}+\frac{k}{3}$$ or $$4u^2-\left(\frac{2k}{3}-\frac{5}{4}\right)u+\frac{k}{3}-\frac{5}{4}\geq0,$$ for which it's enough to prove that $$\left(\frac{2k}{3}-\frac{5}{4}\right)^2-16\left(\frac{k}{3}-\frac{5}{4}\right)\leq0,$$ which is true for $k=\frac{27}{8}+\frac{9\sqrt{3}}{4}.$
Let $b=c=1$.
Thus, we need to prove that: $$\frac{a^3+2}{2(a+1)^2}+\frac{k(2a+1)}{(a+2)^2}\geq\frac{3}{8}+\frac{k}{3}$$ or $$\frac{4a^3-3a^2-6a+5}{8(a+1)^2}\geq\frac{k(a-1)^2}{3(a+2)^2}$$ or $$(4a+5)(a+2)^2\geq(9+6\sqrt3)(a+1)^2$$ or $$(2a+4-\sqrt3)(2a+1-\sqrt3)^2\geq0$$ and we are done!