<s>[INST] $a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$  
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Again, we can have, $ab+bc+ca\le b(3-b)+\dfrac{1}{b}$
Now, I have to show that, $b(3-b)+\dfrac{1}{b}\le 3$
How can I prove this now? Please help.
 [/INST] $$\Longleftrightarrow (a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \le 27a^2b^2c^2$$
Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive).  If $x < 0$ then $$(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$$. 
 If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes
$$64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$$
Note that $$9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$$ and $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$.  Combining these completes our proof!
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