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.

My first SOS is same as Nguyen Huyen. Here is my second and third.
$$\displaystyle a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$
$$\displaystyle=\frac34 \left( b-c \right) ^{2} \left( b+c \right) +\frac34 \left( a-b
 \right) ^{2}a+\frac14\, \left( a+b-2c \right) ^{2}a+\frac12\, \left(b+c-2a \right) ^{2}a$$
$$\displaystyle =\frac34\cdot {\frac { \left( 2\,{a}^{2}-2\,ac+{b}^{2}-{c}^{2} \right) ^{2}}{2
\,a+b+c}}+{\frac {3 a\left( a+b+c \right)\left( a-b \right) ^{2} }{
2\,a+b+c}}$$
