<s>[INST] 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.
 [/INST] Because $$a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$
$$=\frac{3}{4}\Big[4a^3-4(b+c)a^2+2(b^2+c^2)a+b^3-b^2c-bc^2+c^3\Big]$$
$$=\frac{3}{4}\Big[a(2a-b-c)^2+a(b-c)^2+(b+c)(b-c)^2\Big]\geqslant 0.$$
The equality occurs for $$a(2a-b-c)^2=a(b-c)^2=(b+c)(b-c)^2=0.$$

*

*$a=0$.

Thus, $(b+c)(b-c)^2=0,$ which gives $b=c$.


*$a\neq0$.

Thus, since $a(b-c)^2=0,$ we obtain $b=c$ and since $a(2a-b-c)^2=0,$
we obtain $a=b=c$.
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