Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$

Question

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$?$

Answer 1

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.

1. $w^3=0$.

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 $k=\frac{27}{8}+\frac{9\sqrt{3}}{4}.$

2. Two variables are equal.

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!

Answer 2

Another way

Let $p = a + b + c = 3, q = ab+bc+ca, r = abc$.

Since $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ that is $-4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$, we have \begin{align} r &\ge -\frac{2}{27}p^3 + \frac{1}{3}pq - \frac{2}{27}\sqrt{(p^2 - 3q)^3}\\ &= q - 2 - 2\sqrt{(1-q/3)^3}.\tag{1} \end{align}

We have to prove that $$\frac{p^3 - 3pq + 3r}{pq - r} - \frac{3}{8} \ge k\left(\frac{1}{3} - \frac{q}{p^2}\right).$$ From (1), it suffices to prove that $$\frac{p^3 - 3pq + 3(q - 2 - 2\sqrt{(1-q/3)^3})}{pq - (q - 2 - 2\sqrt{(1-q/3)^3})} - \frac{3}{8} \ge k\left(\frac{1}{3} - \frac{q}{p^2}\right)$$ or $$\frac{27(1-q/3)(3 - \sqrt{1-q/3})}{8q + 8 + 8\sqrt{(1-q/3)^3}}\ge \frac{k}{3}(1 - q/3).$$ It suffices to prove that $$\frac{81(3 - \sqrt{1-q/3})}{8q + 8 + 8\sqrt{(1-q/3)^3}}\ge k.$$ With the substitution $q = 3 - \frac{t^2}{3}$ for $0 \le t \le 3$, it suffices to prove that $$\frac{729(9-t)}{8(t+3)(6-t)^2} \ge k.$$ It is easy to prove that $$\frac{729(9-t)}{8(t+3)(6-t)^2} - \frac{27}{8} = \frac{27t^2(9 - t) + 27^2(5 - t)}{8(t+3)(6-t)^2} > 0$$ and $$\left(\frac{729(9-t)}{8(t+3)(6-t)^2} - \frac{27}{8}\right)^2 - \frac{243}{16} = \frac{243(99-6t-t^2)(t^2 - 12t + 9)^2}{64(t+3)^2(6-t)^4} \ge 0$$ which results in $$\frac{729(9-t)}{8(t+3)(6-t)^2} \ge \frac{27}{8} + \frac{9\sqrt{3}}{4}.$$ (Alternatively, we may use derivatives to find the minimum of $\frac{729(9-t)}{8(t+3)(6-t)^2}$ on $[0, 3]$.)

We are done.