<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] We consider four cases for $(3- 2a),(3-2b),(3-2c)$ : The case when all of them are non-negative, the case when exactly one of them is negative, the case when exactly two of them are negative, and the case when all of them are negative.
For the first case when all of them are non-negative, notice that by the AM-GM inequality, 
$$\begin{align}(3 - 2a) + (3 - 2b) &\ge 2\sqrt{(3-2a)(3-2b)}\\
6 - 2a - 2b &\ge 2\sqrt{(3-2a)(3 - 2b)}\\
2(a + b + c) - 2a - 2b &\ge 2\sqrt{(3-2a)(3 - 2b)}\\
c &\ge \sqrt{(3-2a)(3 - 2b)}\end{align}$$
Similarly for $(3 - 2a)(3-2c)$ and $(3 - 2b)(3 - 2c)$, we get $b \ge \sqrt{(3-2a)(3-2c)}$ and $a \ge \sqrt{(3-2b)(3-2c)}$ respectively.
Multiplying all $3$ together, we get :
$$\begin{align}abc &\ge\sqrt{(3-2a)^2(3-2b)^2(3-2c)^2}\\
\implies a^2b^2c^2 &\ge (3-2a)(3-2b)(3-2c)\end{align}$$
Hence the inequality holds for all $(3 - 2a), (3 - 2b), (3 - 2c) \ge 0$, with equality at $a = b = c = 1$.
For the second case when exactly one of them is negative, we assume without a loss of generality that $3 - 2a < 0$. Then,
$$\begin{align}&3 - 2a < 0\\
\implies &a > \frac{3}{2}\\
\implies &3 - b - c > \frac{3}{2}\\
\implies &b + c < \frac{3}{2}\\
\implies &(3 - 2b)(3 - 2c) > 0\\
\implies &(3 - 2a)(3 - 2b)(3 - 2c) < 0\end{align}$$
On the other hand, $a^2b^2c^2 \ge 0$, hence the inequality holds true for the case when exactly one of them is negative.
For the case when exactly two of them is negative, we see that (without a loss of generality) $(3 - 2a), (3 - 2b) < 0 \implies a, b > \frac{3}{2}\implies a + b > 3 \implies a + b + c > 3$. But this is impossible since $a + b + c = 0$, leading to a contradiction.
Similarly, we proof again by contradiction that not all of the three can be negative. If, on the contrary all the three are negative, then $a, b, c > \frac{3}{2} \implies a + b + c > \frac{9}{2}$. But this is a contradiction since $a + b + c = 3$.
Therefore, after considering all the cases, we conclude that $$a^2b^2c^2 \ge (3-2a)(3-2b)(3-2c)$$
My initial solution assumed that all of $(3-2a), (3-2b), (3-2c) \ge 0$. But clearly this is not necessarily true, so I took sometime to write out this new solution, though I'm pretty sure there are other shorter methods.
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