# 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.

• try the substitution $x = \frac{b+c}{2}$, $y = \frac{b-c}{2}$, and work in terms of $a, x, y$ instead of $a, b, c$ Aug 10, 2020 at 5:41

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 $$a(2a-b-c)^2=a(b-c)^2=(b+c)(b-c)^2=0.$$

1. $$a=0$$.

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

1. $$a\neq0$$.

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

we obtain $$a=b=c$$.

• I can see that you "brute-force" most problems, but it is difficult to see equality scenario then. Aug 10, 2020 at 6:42
• @Book Of Flames For $a=b=c$ or $b=c$ and $a=0$ of course. Aug 10, 2020 at 6:56
• How do you know that these are only cases where the equality holds? Is it a guess? Aug 10, 2020 at 7:02
• @BookOfFlames I think it just take $b=c$ (because they are symmetric) then easy see the equality cases. Aug 10, 2020 at 7:04
• @Book Of Flames I added something. See now. Aug 10, 2020 at 8:43

My first SOS is same as Nguyen Huyen. Here is my second and third.

$$\displaystyle a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$

$$\displaystyle=\frac34 \left( b-c \right) ^{2} \left( b+c \right) +\frac34 \left( a-b \right) ^{2}a+\frac14\, \left( a+b-2c \right) ^{2}a+\frac12\, \left(b+c-2a \right) ^{2}a$$ $$\displaystyle =\frac34\cdot {\frac { \left( 2\,{a}^{2}-2\,ac+{b}^{2}-{c}^{2} \right) ^{2}}{2 \,a+b+c}}+{\frac {3 a\left( a+b+c \right)\left( a-b \right) ^{2} }{ 2\,a+b+c}}$$

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 $$a^3+b^3+c^3-3abc-{2\left(\frac{b+c}{2}-a\right)^3}$$ $$=(b-c)^2+\frac{3\,a}{2} \Big[(a-b)^2+(b-c)^2+(c-a)^2\Big] \geq 0.$$

• My SOS is same. Aug 10, 2020 at 6:24

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 $$F=0\,\, \forall a >0$$