<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] 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
$$(-a+b+c)(a-b+c)(a+b-c)\le(b+c-b)(a+c-c)(a+b-a)=abc$$
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