Let $a,b,c$ be non-negative reals, and $k$ is the best possible constant Prove the inequality Let $a,b,c$ be nonnegative  real numbers. Prove that
$$a^3+b^3+c^3-3abc\geq k|(a-b)(b-c)(c-a)|$$
where $k=\left(\frac{27}{4}\right)^{1/4}(1+\sqrt{3})$ and that $k$ is the best possible constant.
Not sure how to go about proving this inequality, I have not dealt with a type of inequality that includes a best possible constant $k$.
Any help to solve and understand this would be greatly appreciated.

The problem amounts to minimizing:
$$ R(a,b,c)= \frac{a^3+b^3+c^3 - 3 a bc }{(b-a)(c-b)(c-a)} $$
on the set $0\leq a <b< c$.
The directional derivative:
$$ \frac{d}{dt}_{|t=0} R(a+t,b+t,c+t)= \frac{3}{2}\frac{(b-a)^2+(c-b)^2+(c-a)^2 }{(b-a)(c-b)(c-a)} $$ is non-negative so any minimum must occur for $a=0$. By homogeneity we may assume $c=1$. So we are reduced to minimizing for  $0<b<1$:
$$ R(0,b,1)=\frac{1+b^3}{(1-b)b} .$$
And this happens (set derivative=0)  when 
$$0=b^4-2b^3-2b+1=(b^2-(1+\sqrt{3})b+1)(b^2-(1-\sqrt{3})b+1)$$
Only the first factor has real roots from which:
$$ b=\frac{1+\sqrt{3}}{2}- \sqrt{\frac{\sqrt{3}}{2}} \ \ \left( \ \mbox{and}
\ \ \ \frac{1}{b}=\frac{1+\sqrt{3}}{2}+ \sqrt{\frac{\sqrt{3}}{2}} \right)$$
Inserting into the previous it yields (after reduction) the cited constant $k$. To see this note that:
 $$ (1-b) \left[ (1-b)(1+b^2) \right] =(1-b)^2(1+b^2) = 1-2b+2b^2-2b^3+b^4 = 2 b^2$$
(use that $b$ is the root of the 4'th degree polynomial above).
Using also $1+b^3=(1+b)(1-b+b^2)=(1+b) \sqrt{3} b$ we get for the minimal value:
$$  \frac{1+b^3}{b(1-b)} =  \frac{(1-b)(1+b^2) (1+b)\sqrt{3}}{2 b^2}=\frac{\sqrt{3}}{2}\left( \frac{1}{b}+b\right) \left( \frac{1}{b} - b\right)=\sqrt{3} \left( 1+ \sqrt{3}\right) \sqrt{\frac{\sqrt{3}}{2}}$$
