Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers 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.

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.
