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.

Using algebra.
Consider that we look for the minimum value of
$$F=a^3+b^3+c^3-3abc -2\left({b+c\over 2}-a\right)^3$$
$$\frac{\partial F}{\partial a}=3 a^2+6 \left(\frac{b+c}{2}-a\right)^2-3 b c \tag 1$$
$$\frac{\partial F}{\partial b}=3b^2-3 \left(\frac{b+c}{2}-a\right)^2-3 a c\tag 2$$
$$\frac{\partial F}{\partial c}=3c^2-3 \left(\frac{b+c}{2}-a\right)^2-3 a b\tag 3$$
Using $(3)$ the only possible value of $c$ is
$$c=\frac{1}{3} \left(2 \sqrt{4 a^2-a b+b^2}-2 a+b\right)\tag 4$$ Plug in $(2)$ and solve for $b$; the only possible solutions are
$$b_1=a \qquad \text{and} \qquad b_2=\frac{\sqrt{6}-1}{2}  a\tag 5$$
$$b=b_1\implies\frac{\partial F}{\partial a}=0$$
$$b=b_2\implies\frac{\partial F}{\partial a}=\frac{1}{12} \left(359-140 \sqrt{6}\right) a^2$$ which must be rejected.
So $b=a$ and $c=a$ then $a=b=c$ and $F=0\,\, \forall a >0$
