<s>[INST] Prove that $\dfrac{x^3+y^3+z^3}{3} \geq xyz + \dfrac{3}{4}|(x-y)(y-z)(z-x)|$ 
For nonnegative real numbers $x,y,$ and $z$ prove that $$\dfrac{x^3+y^3+z^3}{3} \geq xyz + \dfrac{3}{4}|(x-y)(y-z)(z-x)|.$$

The absolute value sign seems to make this difficult. Should I prove this in two cases? One for $(x-y)(y-z)(z-x) \geq 0$ and one for $(x-y)(y-z)(z-x) < 0$?
 [/INST] We can prove a stronger inequality:
$$\frac{x^3+y^3+z^3}{3} \geq xyz + |(x-y)(y-z)(z-x)|$$
Using this equality:
$$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2)+(z-x)^2]$$
we have:
$$(x+y+z)[(x-y)^2+(y-z)^2)+(z-x)^2] \geq 6|(x-y)(y-z)(z-x)|$$
Assume $x \leq y \leq z$. Let: $y-x = a, z-y = b, z-x = c$, then $a+b=c$ and $a,b,c \geq 0$
The inequality becomes:
$$(x+y+z)(a^2+b^2+c^2) \geq 6abc$$
This is true because:
$$x+y+z \geq z-x = c \geq 0$$
$$a^2+b^2+c^2 = a^2+b^2+(a+b)^2 \geq 6ab \geq 0$$
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