Problem with proving inequalities Question:

Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.

My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {xyz}$$ Cubing both sides and multiplying by $8$, $$\frac{8a^3}{27}>8xyz$$
Also, by $AM>GM$, $$(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)>8xyz$$
Now, how do I find the relation between $(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)$ and $\frac{8a^3}{27}$?

Use AM-GM of three items as
$$F=[(a-x)(a-y)(a-z)]^{1/3} \le \frac{3a-(x+y+z)}{3}$$
$$\implies (a-x)(a-y)(a-z) \le \frac{8a^3}{27}$$
