SunnyEvan wrote:

Let $a,b,c \in R$ ,such that $a^2+b^2+c^2=4(ab+bc+ca)$ Prove that : $\frac{53}{2}-9\sqrt{14} \leq \frac{8(a^3b+b^3c+c^3a)}{27(a^2+b^2+c^2)^2} \leq \frac{53}{2}+9\sqrt{14}$

We have the identity
$a^3b + b^3c + c^3a = (a+b+c)(a^2b + b^2c + c^2a) - abc(a+b+c) - (a^2b^2 + b^2c^2 + c^2a^2)$ $\Rightarrow a^3b + b^3c + c^3a = e_1(a^2b + b^2c + c^2a) + e_1e_3 - e_2^2$
WLOG $a^2 + b^2 + c^2 = 1$

Enough to show that
$\frac{1432 - 486\sqrt{14}}{16} \geq \sqrt{\frac{3}{2}} (a^2b + b^2c + c^2a + abc)$
We have,
$abc = (a+b+c)(ab+bc+ca) - (a+b)(b+c)(c+a) = \sqrt{\frac{3}{2}}\left(\frac{1}{4}\right) - (a+b)(b+c)(c+a)$
Also,
$a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2 = (a+b)(b+c)(c+a) - 2abc = -(a+b)(b+c)(c+a) - 2\sqrt{\frac{3}{2}}\left(\frac{1}{4}\right)$
Let $(a+b)(b+c)(c+a) = x$

Hence the inequality becomes $x \geq -10$ (approx). But we have
$x \geq 8(abc)^{2/3}$
Hence it is enough to show that
$(abc)^{2/3} \geq -\frac{5}{4} \Rightarrow \text{True since } (abc)^{2/3} \geq 0 \text{ for real } a,b,c$