Proving $\frac{a+b+c}{a^2b+b^2c+c^2a+9}\ge \frac{abc+6}{abc+27}$ for $a, b, c\ge 0$; $ab + bc + ca = 3$

Question

I came up with the inequality accidentally so there is no original proof so far. It would be great if you can give some useful help to prove it.

Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=3.$ Prove that$$\frac{a+b+c}{a^2b+b^2c+c^2a+9}\ge \frac{abc+6}{abc+27}.$$ Equality occurs when $a=b=c=1$ or $a=3;b=1;c=0$ and its permutations.

I thought that $BW$ might be helpful for this case. See more about this method AOPS.

Firstly, I tried to verify the inequality as $$abc(a+b+c)+27(a+b+c)\ge abc(a^2b+b^2c+c^2a)+9abc+54+6(a^2b+b^2c+c^2a).$$ After homogenization, it's

$$\frac{abc(a+b+c)(ab+bc+ca)}{3}+\sqrt{3}(a+b+c)\sqrt{(ab+bc+ca)^5}$$ $$\ge abc(a^2b+b^2c+c^2a)+\sqrt{3}abc\sqrt{(ab+bc+ca)^3}+2(ab+bc+ca)^3+\frac{2\sqrt{3}(a^2b+b^2c+c^2a)\sqrt{(ab+bc+ca)^3}}{3}.$$

Or $$3 abc(a^2b+b^2c+c^2a)+6(ab+bc+ca)^3-abc(a+b+c)(ab+bc+ca)\le \sqrt{3(ab+bc+ca)^3}\cdot\left[3(a+b+c)(ab+bc+ca)-3abc-2(a^2b+b^2c+c^2a)\right].$$

It's not hard to see that both side is non-negative sign.

Hence, it's enough to show a $12$th degree cyclic polynomial

$$3(ab+bc+ca)^3\cdot\left[3(a+b+c)(ab+bc+ca)-3abc-2(a^2b+b^2c+c^2a)\right]^2\ge \left[3 abc(a^2b+b^2c+c^2a)+6(ab+bc+ca)^3-abc(a+b+c)(ab+bc+ca)\right]^2.$$ I attach the result by Wolfram. BW

I did not check all coefficient. In case it is true, the proof is still non-human proof, I think.

Also, I found an interesting fact which is simpler $$\boxed{\color{green}{\frac{9(a+b+c)}{2}-\frac{3abc}{2}-9\ge a^2b+b^2c+c^2a}.}$$ This inequality saves desired occuring equality.

I try to use it for the starting inequality without success.

Indeed, we'll prove $$\color{black}{\frac{a+b+c}{\dfrac{9(a+b+c)}{2}-\dfrac{3abc}{2}}\ge \frac{abc+6}{abc+27}}$$which is already wrong when $a=b\rightarrow 1^{-}.$

For this kind of inequalities, I'd like to study more about the relations of $a^2b+b^2c+c^2a, a+b+c,ab+bc+ca, abc.$

It is very useful if you can give some relevant links to topic.

I truly hope we can find some nice ideas for my inequality. Thank you for sharing.

Answer 1

Here is a proof.

If $abc = 0$, it is easy.

In the following, assume that $abc > 0$.

The desired inequality is written as $$\frac{(a + b + c)(abc + 27)}{abc + 6} - 9 - (a^2b + b^2c + c^2a) \ge 0$$ or \begin{align*} &\frac{(a + b + c)(abc + 27)}{abc + 6} - 9 - \frac{a^2b + b^2c + c^2a + (ab^2 + bc^2 + ca^2)}{2} \\[6pt] \ge{}& \frac{a^2b + b^2c + c^2a - (ab^2 + bc^2 + ca^2)}{2}\\[6pt] ={}& \frac{(a - b)(b - c)(a - c)}{2}. \tag{1} \end{align*}

It is easy to prove that $$\frac{(a + b + c)(abc + 27)}{abc + 6} - 9 - \frac{a^2b + b^2c + c^2a + (ab^2 + bc^2 + ca^2)}{2} \ge 0. \tag{2}$$ (One may use pqr method.)

From (1) and (2), it suffices to prove that \begin{align*} &\left(\frac{(a + b + c)(abc + 27)}{abc + 6} - 9 - \frac{a^2b + b^2c + c^2a + (ab^2 + bc^2 + ca^2)}{2} \right)^2\\[6pt] \ge{}& \frac{(a - b)^2(b - c)^2(a - c)^2}{4}. \tag{3} \end{align*}

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca = 3, r = abc$. (3) is written as \begin{align*} f(p) &:= \left( {r}^{3}+12\,{r}^{2}+36\,r \right) {p}^{3}+ \left( -2\,{r}^{2}- 45\,r+243 \right) {p}^{2}\\ &\qquad + \left( -15\,{r}^{3}-108\,{r}^{2}-432\,r- 1944 \right) p\\ &\qquad +9\,{r}^{4}+81\,{r}^{3}+108\,{r}^{2}+324\,r+3888 \\ &\ge 0. \tag{4} \end{align*}

If $r = 1$, we have $f(p) = 49(p+10)(p-3)^2 \ge 0$.

If $r \ne 1$, the discriminant of $f$ is give by $-27(r + 6)^2(r - 1)^2r g(r) < 0$ where \begin{align*} g(r) &:= 81\,{r}^{9}+2592\,{r}^{8}+33340\,{r}^{7}+224198\,{r}^{6}+951556\,{r}^{ 5} +3657420\,{r}^{4}\\ &\qquad +14070429\,{r}^{3}+34706232\,{r}^{2}+39602196\,r+ 10628820. \end{align*} Thus, the cubic equation $f(p)= 0$ has exactly one real root on $(-\infty, \infty)$. Note that $f(0) > 0$ and $f(-\infty) = -\infty$. Thus, $f(p) = 0$ has a negative real root. Thus, $f(p) \ge 0$ for all $p \ge 0$.

We are done.