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 $(x-y)(y-z)(z-x) < 0$?

First from AM-GM inequality,
$$
\left(|(x-y)(y-z)(z-x)|\right)^{1/3}
\leq \frac{|x-y|+|y-z|+|z-x|}{3}\leq \frac{2}{3}(x+y+z)
$$
and
\begin{align}
\left(|(x-y)(y-z)(z-x)|\right)^{2/3}
&=\left(|(x-y)^2(y-z)^2(z-x)^2|\right)^{1/3}\cr
&\leq \frac{(x-y)^2+(y-z)^2+(z-x)^2}{3}\cr
&=\frac{2}{3}(x^2+y^2+z^2-xy-yz-xz).
\end{align}
Therefore,
\begin{align}
\frac{x^3+y^3+z^3}{3}-xyz&=\frac{1}{3}(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\cr
&\geq \frac{3}{4}|(x-y)(y-z)(z-x)|.
\end{align}
