$a+b+c=3, a,b,c>0$, 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.

Here's a geometric method, just for variety.
First (as in Yiyuan Lee's solution) we consider cases: how many of the factors on the RHS are negative?
Case 3: All three are negative; then the RHS is negative and we are done.
Case 2: Exactly two are negative; wlog, $3-2a<0$ and $3-2b<0$.  This is impossible because $3-2a = a+b+c-2a = -a+b+c$, so $3-2a<0$ means $b+c<a$; similarly $3-2b<0$ means $a+c<b$; adding these inequalities yields $c<0$.
Case 1: Exactly one is negative; same as case 3.
Case 0: All three are positive.  Then, as in case 2, we obtain $b+c>a$ and $a+c>b$ and $a+b>c$.  Therefore there is a triangle with sides $a$,$b$,$c$.  Using a standard formula for the circumradius $R$ and our hypothesis that $a+b+c=3$, we get
$$ R = \frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}
= \frac{abc}{\sqrt{3(3-2a)(3-2b)(3-2c)}} $$
and so the desired inequality is equivalent to $R\ge\frac1{\sqrt3}$.  So we want to prove this:

Any triangle with perimeter $3$ has circumradius at least $\frac1{\sqrt3}$.  

Equivalently:

Any triangle with circumradius $1$ has perimeter at most $3\sqrt3$.

Equivalently:

Of all triangles with circumradius $1$, the equilateral triangle has the greatest perimeter.

By a standard compactness argument, there exists a triangle with maximum perimeter among those with circumradius $1$; to show that any such triangle is equilateral triangle, apply the following lemma to each vertex.
Lemma. Let $P$ and $Q$ be fixed points on a circle, and let $X$ be a variable point on that circle.  If $PX+XQ$ is maximized then $X$ lies on the diameter bisecting the chord $PQ$.
Proof of lemma.  Let $XY$ bisect $\angle PXQ$, with $Y$ on $PQ$.  Drop perpendiculars from $P$ and $Q$ to $XY$, with feet $P'$ and $Q'$ respectively.  

Then
$$ PX+XQ = \frac{PP'+Q'Q}{\sin(\frac12\angle PXQ)}
\le \frac{PY+YQ}{\sin(\frac12\angle PXQ)}
= \frac{PQ}{\sin(\frac12\angle PXQ)}
$$
Note that $\angle PXQ$ does not depend on the position of $X$ (Euclid III:21), only on what side of $PQ$ it falls on*, and that we have equality exactly when $Y$, $P'$, and $Q'$ coincide, that is, $XY$ is perpendicular to $PQ$, as desired.
*It doesn't matter for us which of the two cases is the sharp one, but of course it's when $X$ is on the major arc associated with $PQ$.
