Norbert's answer explains the case when $a,b,c$ are not the sides of a triangle.
Let $x,y,z$ be as in this picture, where $AB=c, BC=a, CA=b$.
Then $(a,b,c)=(y+z,x+z,x+y)$.$\,$ Inequality becomes $$8xyz\le (x+y)(y+z)(z+x),$$
which as Norbert says is true by $2\sqrt{xy}\le x+y$ (proof: $\Leftrightarrow (\sqrt{x}-\sqrt{y})^2\ge 0$) and $2\sqrt{yz}\le y+z$ and $2\sqrt{zx}\le z+x$.
Some overkill approaches: by Hölder's inequality: $$(x+y)(y+z)(z+x)\ge (\sqrt[3]{xyz}+\sqrt[3]{yzx})^3=8xyz$$
By AM-GM: $$(x+y)(y+z)(z+x)-xyz=(x+y+z)(xy+yz+zx)$$ $$\ge (3\sqrt[3]{xyz})(3\sqrt[3]{xy\cdot yz\cdot zx})=9xyz$$
By Cauchy-Schwarz: $$(z+x+y)(xy+yz+zx)\ge (\sqrt{z}\cdot \sqrt{xy} + \sqrt{x}\cdot \sqrt{yz}+\sqrt{y}\cdot \sqrt{zx})^2=9xyz$$
By Muirhead's inequality:
$$(x+y)(y+z)(z+x)-2xyz=\sum_{\text{sym}}x^2y^1z^0\ge \sum_{\text{sym}}x^1y^1z^1=6xyz,$$
because $(2,1,0)\succ (1,1,1)$. Last one is also true by rearrangement inequality because hint: we can let WLOG $x\ge y\ge z$ and $xy \ge zx\ge yz$.