show that $a^3+b^3+c^3-3abc\ge2(\frac{b+c}{2}-a)^3$

Question

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

Answer 1

First of all the inequality is trivial for $b+c-2a<0$. Secondly, you have: $$ 2(a^2+b^2+c^2-ab-bc-ac)=(a-b)^2+(c-b)^2+(a-c)^2 $$ Also you have: $$ (a-b)^2+(a-c)^2\geq 2(b-a)(c-a) $$ and therefore: $$ 2(a-b)^2+2(a-c)^2+2(c-b)^2\geq (a-b)^2+(a-c)^2+2(b-a)(c-a) =(b+c-2a)^2 $$ hence: $$ (a^2+b^2+c^2-ab-bc-ac)\geq (\frac{b+c-2a}{2})^2 $$ But you also have $a+b+c\geq b+c-2a$. Now if you multiply this by the previous inequality you get the answer.

Answer 2

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 $(0, t, t)$