Problem with proving inequalities

Question

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 $\frac{8a^3}{27}$?

Answer 1

You can proceed like this: $$(a-x)(a-y)(a-z) \leqslant \left( \dfrac{(a-x)+(a-y)+(a-z)}{3} \right) ^3 =\dfrac{8}{27} a^3$$

Answer 2

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

Answer 3

Using AM-GM identity,

$\implies$ $[(a-x)(a-y)(a-z)]^{1/3} \le \frac{(a-x)+(a-y)+(a-z)}{3}$$

Then we end up with our desired result

$$\implies (a-x)(a-y)(a-z) \le \frac{8a^3}{27}$$

Answer 4

Let $x=y\rightarrow0^+$ and $z\rightarrow a$.

Thus, the left side is closed to $0$, but the right side is greater than $0$,

which says that our inequality is not true.