<s>[INST] 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.
 [/INST] 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.
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