Define $3x=a+b+c$, $3y^2=ab+bc+ca$, $z^3=abc$. Expand the LHS to find that it is equivalent to $$\frac{a^2+b^2+c^2+3(ab+bc+ca)}{(a+b+c)(ab+bc+ca)-abc}=\frac{9x^2+3y^2}{9x^2y-z^3}.$$ Use $sr=A=\frac{abc}{4R}$ and substitute stuff for $R$ to find that $\frac{s}{16Rr}=\frac{s^2}{4abc}=\frac{9x^2}{16z^3}$. We also find $\frac{r}{16Rs}=\frac{r^2}{4abc}=\frac{A^2}{4abcs^2}=\frac{A^2}{9x^2z^3}=\frac{16A^2}{144x^2z^3}$, and since $16A^2=2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4$ (by Heron's) we find after simplifying that $\frac{16A^2}{144x^2z^3}=\frac{3y^2-\frac{9}{4}x^2}{4z^3}-\frac{1}{6x}$. Finally, $\frac{11}{8s}=\frac{11}{12x}$. Thus the inequality can be written as $$\frac{9x^2+3y^2}{9x^2y-z^3} \leqslant \frac{11}{12x}+\frac{9x^2}{16z^3}+\frac{3y^2-\frac{9}{4}x^2}{4z^3} - \frac{1}{6x} = \frac{3}{4x} + \frac{3y^2}{4z^3},$$which after cross-multiplying gives $$12x^3z^3+5xy^2z^3+z^6 \leqslant 9x^2yz^3 + 9x^3y^3.$$ Note that since $x \geqslant y \geqslant z$, every term on the RHS is bigger than any term on the LHS, except for $x^3z^3$ on the left and $x^2yz^3$ on the right. Thus it suffices to show $9x^3y^3+3x^2yz^3 \geqslant 12x^3z^3$. Since this is homogeneous, assume $y=1$ so we wish to show $3x+z^3 \geq 4xz^3$. Write this as $3x \geqslant z^3(4x-1)$. We also know (say, by Muirhead) that $xz^3\leqslant y^4 = 1$, so for a fixed value of $x$, the LHS is fixed, and the RHS is maximal when $z^3$ is maximal, so we only need to consider the case when $z^3=\frac{1}{x}$. In this case we want to show that $3x \geqslant 4 -\frac{1}{x}$, or $3x^2+1 \geqslant 4x$, or $(x-1)(3x-1)\geqslant 0$, which is true since $x\geqslant y=1$. This concludes the proof. Equality happens when $x=1$, but $y=1$, so $x=y$, which gives $a=b=c$. Thus equality holds only in an equilateral triangle. It would be great if someone could find a nice proof for this problem.