<s>[INST] 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 
 [/INST] By AM-GM, we have $a^3 + b^3 + c^3 \ge 3abc$.  So if $\dfrac{b+c}{2}-a \le 0$ the inequality is done.  
So let us assume $\dfrac{b+c}{2}-a > 0$, and we can define $b = a + 2x, c = a+2y$, with $x+y > 0$.  Then,
$$a^3 + b^3 + c^3 - 3abc - 2\left(\dfrac{b+c}{2}-a\right)^3 \\
 = 12a(x^2-xy+y^2) + 6(x+y)(x-y)^2 \ge 0
$$
Equality is when $(a, b, c) = (t, t, t)$ or $(0, t, t)$
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