I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
Case 1. If $a,b,c$ are lengths of triangle.
Since $$ 2\sqrt{xy}\leq x+y\qquad 2\sqrt{yz}\leq y+z\qquad 2\sqrt{zx}\leq z+x $$ for $x,y,z\geq 0$, then multiplying this inequalities we get $$ 8xyz\leq(x+y)(y+z)(z+x) $$ Now substitute $$ x=\frac{a+b-c}{2}\qquad y=\frac{a-b+c}{2}\qquad z=\frac{-a+b+c}{2}\qquad $$ Since $a,b,c$ are lengths of triangle, then $x,y,z\geq 0$ and our substitution is valid. Then we will obtain $$ (-a+b+c)(a-b+c)(a+b-c)\leq abc\tag{1} $$
Case 2. If $a,b,c$ are not lengths of triangle.
Then at least one factor in left hand side of inequality $(1)$ is negative. In fact the only one factor is negative. Indeed, without loss of generality assume that $a+b-c<0$ and $a-b+c<0$, then $a=0.5((a+b-c)+(a-b+c))<0$. Contradiction, hence the only one factor is negative. As the consequence left hand side of inequality $(1)$ is negative and right hand side is positive, so $(1)$ obviously holds.
Here's a geometric proof:
If $a,b,c$ satisfy the triangle inequality, let $A$ be the area of the triangle $T$ with side lengths $a$, $b$, $c$. Then the inequality reduces to $$\frac{16A^2}{a+b+c} \leq abc$$ by Heron's formula. Since $A$ is positive, this is equivalent to $$\frac{2A}{a+b+c} \leq \frac{abc}{8A} \, .$$
But the left-hand side of this last inequality is the inradius of $T$ while the right-hand side is the radius of $T$'s nine-point circle. Hence the inequality follows from Feuerbach's theorem (EDIT: or from the much simpler and more elementary argument given here).
If $a,b,c$ do not satisfy the triangle inequality, then the inequality is trivial as previously noted.
Here I give a detailed proof. Though steps could have been jumped to keep it short.
Without loss of generality we can assume that $a\ge b \ge c$
Let $$(-a+b+c)(a-b+c)(a+b-c)=S$$ $$\Rightarrow S=(-a+b+c)\{a-(b-c)\}\{a+(b-c)\}$$ $$\Rightarrow S=(-a+b+c)\{a^2-(b-c)^2\} $$ $$\Rightarrow S= (-a+b+c)\{a^2-b^2-c^2+2bc\}$$ $$\Rightarrow S=-(a^3+b^3+c^3)-2abc+b^2c+bc^2+ab^2+a^2b+ac^2+a^2c $$ $$\Rightarrow abc-S=(a^3+b^3+c^3)+3abc-(b^2c+bc^2+ab^2+a^2b+ac^2+a^2c) $$ $$\Rightarrow abc-S=(a^3-a^2b)+(b^3-b^2c)+(c^3-c^2a)+(abc-bc^2)+(abc-ab^2)+(abc-a^2c) $$$$\Rightarrow abc-S=a^2(a-b)+b^2(b-c)+c^2(c-a)+bc(a-c)+ab(c-b)+ac(b-a) $$ $$\Rightarrow abc-S=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)\{a(a-c)-b(b-c)\}+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)^2\{a^2-b^2+c(b-a)\}+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)^2\{a+b-c\}+c(c-a)(c-b)$$ Now $(c-a) \le 0$ and $(c-b) \le 0$ $$\Rightarrow c(c-a)(c-b)\ge 0 $$ and $$ (a-b)^2(a+b-c) \ge 0$$ This shows $$ abc-S \ge 0$$ $$\Rightarrow abc\ge S$$ $$\Rightarrow abc\ge (-a+b+c)(a-b+c)(a+b-c) $$
Note that no two of $(-a+b+c)$, $(a-b+c)$, and $(a+b-c)$ can be negative. If so, then one of $$ \begin{align} (a-b+c)+(a+b-c)&=2a\\ (a+b-c)+(-a+b+c)&=2b\\ (-a+b+c)+(a-b+c)&=2c \end{align} $$ would be negative, but each of $a$, $b$, and $c$ is positive. Thus, at most one can be negative. If only one were negative, then the product on the left would be non-positive and the inequality would be trivial. Therefore, we can assume that $a$, $b$, and $c$ are sides of a triangle.
By Heron's Formula, a triangle with sides of length $a$, $b$, and $c$, has area $A$ where $$ (a+b+c)(-a+b+c)(a-b+c)(a+b-c)=16A^2\tag{1} $$ Let $r$ be the radius of the inscribed circle and $R$ be the radius of the circumscribed circle. Then $$ 2A=r(a+b+c)\tag{2} $$ and $$ 4AR=abc\tag{3} $$ Putting together $(1)$-$(3)$ yields $$ (-a+b+c)(a-b+c)(a+b-c)R=2rabc\tag{4} $$ The distance $d$ between the incenter and circumcenter is given by $$ d^2=R(R-2r)\tag{5} $$ Combining $(4)$ and $(5)$ gives $$ \begin{align} (-a+b+c)(a-b+c)(a+b-c) &=\left(1-\frac{d^2}{R^2}\right)abc\\[6pt] &\le abc\tag{6} \end{align} $$
Justifications of $(2)$, $(3)$, and $(5)$ can be found in this answer.
The inequality is equivalent to
$\displaystyle \sum_{\text{cyc}} a^3 + 3abc \ge \displaystyle \sum_{\text{sym}} a^2b$,
which follows directly from Schur's Inequality.
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: $$(z+x+y)(xy+yz+zx)\ge (\sqrt{z}\cdot \sqrt{xy} + \sqrt{x}\cdot \sqrt{yz}+\sqrt{y}\cdot \sqrt{zx})^2=9xyz$$
$$(x+y)(y+z)(z+x)-2xyz=\sum_{\text{sym}}x^2y^1z^0\ge \sum_{\text{sym}}x^1y^1z^1=6xyz,$$
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 $xy \ge zx\ge yz$.
A proof using the order inequality:
As @robjon points out, no two of $-a+b+c, a-b+c, a+b-c$ can be negative.
If exactly one of them is negative, the inequality is trivial.
Hence, assume that $-a+b+c, a-b+c, a+b-c > 0$.
Now, it's not difficult to see that $(b+c, a+c, a+b)$ and $(-a, -b, -c)$ are equally ordered.
And as $-a+b+c, a-b+c, a+b-c > 0$, it follows from the order inequality for products that
$$(-a+b+c)(a-b+c)(a+b-c)\le(b+c-b)(a+c-c)(a+b-a)=abc$$
