<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.


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

 [/INST] Let $a\geq b\geq c$.
Hence, $$abc-(a+b-c)(a+c-b)(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$
$$=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=$$
$$=(a-b)(a(a-c)-b(b-c))\geq0$$
because $a\geq b\geq0$ and $a-c\geq b-c\geq0$.
Done!
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