<s>[INST] The inequality $ abc\geq(a-b+c)(a+b-c)(b+c-a)$ holds for $a,b,c\geq 0$ 
What is the proof that:
  $$\forall \ a,b,c\geq 0:\quad a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$

I tried :
we can write that expression $(a-b+c)(a+b-c)(b+c-a)$ as $$-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
then
$$a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
$$\Longleftrightarrow $$
$$a b c \geq -a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
Now consider  $f(a)=abc-(-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3)$ for $b,c$ fixed
if i follow the metode of Quang Hoang
let $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes
$$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$
for $x,y,z \geq 0$
we use AM-GM inequality 
$$(y+z)(z+x)(x+y)\ge 2(yz)^{1/2}(2zx)^{1/2}(2xy)^{1/2}.\tag{1}$$
 but other case  of $x,y,z$  is not clear.


*

*i will be appreciated if  someone  could explain it with more detail

 [/INST] Write $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes
$$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$
Note that at most one of $x,y,z$ can be negative (consider the largest of $a,b,c$). 
Edit: Without loss of generality, assume that $a=\max(a,b,c)$, then $$2x=(a-c)+b\ge b\ge 0, 2z=(a-b)+c\ge c\ge 0.$$
So there are two cases


*

*If $y\le0$: (1) is clear since the LHS is $\ge 0$, the RHS is $\le 0$.

*If $y\ge 0$, then all $x,y,z$ are $\ge 0$. Using AM-GM:
$$x+y\ge 2\sqrt{xy},y+z\ge 2\sqrt{yz},z+x\ge 2\sqrt{zx}.$$
Multiplying those gives (1).

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