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.

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 $$ abc-S \ge 0$$ $$\Rightarrow abc\ge S$$ $$\Rightarrow abc\ge (-a+b+c)(a-b+c)(a+b-c) $$ 
