<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] I think you can check this inequality by Lagrange multiplier method. Consider the following Lagrange function:
\begin{equation}
L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3)
\end{equation}
For finding the extreme value, we have:
\begin{equation}
\frac{\partial L}{\partial a}=2ab^2c^2+2(3-2b)(3-2c)+\lambda=0\\
\frac{\partial L}{\partial b}=2a^2bc^2+2(3-2a)(3-2c)+\lambda=0\\
\frac{\partial L}{\partial c}=2a^2b^2c+2(3-2a)(3-2b)+\lambda=0\\
\frac{\partial L}{\partial \lambda}=a+b+c-3=0
\end{equation}
Because $a,b,c$ is symmetric in systems, I only show part of the solutions and the corresponding extreme values:
\begin{equation}
a=1,b=1,c=1,L=0\\
a=\frac{3}{2},b=\frac{3}{2},c=0,L=0\\
a=\sqrt[3]{2},b=\sqrt[3]{2},c=3-2\sqrt[3]{2},L\approx 0.1107011764
\end{equation}
So, we obtain the minimum of $L(a,b,c)$ is $0$. In summary, under the constrain $a+b+c=3$, we have $L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)\geq0$.
This is the inequality you want to prove.
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