On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
$$ \Rightarrow 2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq (b+c-2a)^3 $$
But couldn't take it any further, but it looks as if Hölder's Inequality may help. And also I don't think that it is a "warm-up" problem.
Any help will be appreciated.

If $b+c<2a$ then $$LHS \geq 0 \ge RHS.$$
If $b+c \geq 2a.$ We write inequality as
$$4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3,$$
or
$$4(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \geq  (b+c-2a)^3.$$
Because $a+b+c = 3a+(b+c-2a) \geq b+c-2a,$ so we will show that
$$4(a^2+b^2+c^2-ab-bc-ca)\geq (b+c-2a)^2,$$
equivalent to
$$3(b-c)^2 \geq 0.$$
Which is true. Equality holds when $ a=b=c$ or $ a=0,\,b=c.$
Note. We have
$$a^3+b^3+c^3-3abc-{2\left(\frac{b+c}{2}-a\right)^3}$$
$$=(b-c)^2+\frac{3\,a}{2} \Big[(a-b)^2+(b-c)^2+(c-a)^2\Big] \geq 0.$$
