show that $a^3+b^3+c^3-3abc\ge2(\frac{b+c}{2}-a)^3$ let $a,b,c\ge 0$,show that: $$a^3+b^3+c^3-3abc\ge2 \left(\dfrac{b+c}{2}-a\right)^3$$
my try:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
then let $b-a=x,c-a=y$
But following I don't can't prove it,Thank you 

First of all the inequality is trivial for $b+c-2a<0$. 
Secondly, you have:
$$
2(a^2+b^2+c^2-ab-bc-ac)=(a-b)^2+(c-b)^2+(a-c)^2
$$
Also you have:
$$
(a-b)^2+(a-c)^2\geq 2(b-a)(c-a) 
$$
and therefore:
$$
2(a-b)^2+2(a-c)^2+2(c-b)^2\geq (a-b)^2+(a-c)^2+2(b-a)(c-a) =(b+c-2a)^2
$$
hence:
$$
(a^2+b^2+c^2-ab-bc-ac)\geq (\frac{b+c-2a}{2})^2
$$
But you also have $a+b+c\geq b+c-2a$. Now if you multiply this by the previous inequality you get the answer.
