By using Taiwanese Transformation, that is, let $ a=x\cdot bc^2,b=y\cdot ca^2, c=z\cdot ab^2 $. Then $ (xyz)\left(a^2 b^2 c^2\right)=1 $, and $ x,y,z \ge 1 $ Notice that we have $ 4(k-1) \le k^2 $ when $ k \ge 1 $ $ E=a^4 b^4 c^4 (x-1)(y-1)(z-1)=\frac{(x-1)(y-1)(z-1)}{x^2 y^2 z^2} \le \frac{1}{64} $.

$bc^2(a-bc^2)\leq\left(\frac{bc^2+a-bc^2}{2}\right)^2=\frac{a^2}{4}\implies \frac{bc^2(a-bc^2)}{a^2}\leq\frac{1}{4},$ $abc(a-bc^2)(b-ca^2)(c-ab^2)=\frac{bc^2(a-bc^2)}{a^2}\cdot\frac{ca^2(b-ca^2)}{b^2}\cdot\frac{ab^2(c-ab^2)}{c^2}\leq\left(\frac{1}{4}\right)^3=\frac{1}{64}.$

Another simple way. We have $E=(ab-b^2c^2)(bc-c^2a^2)(ca-a^2b^2)=(x-y^2)(y-z^2)(z-x^2)$ where $x=ab,y=bc,z=ca$. All the factors are positive. Thus we can apply AM-GM.We do it as follows:$\sqrt[3]{E}\leq \frac{(x-x^2)+(y-y^2)+(z-z^2)}{3}\overset{\bf \color{red} (1)}\leq \frac{1}{4}$ thus $E\leq \left(\frac{1}{4}\right)^3$ with equality when $a=b=c=\frac{1}{\sqrt{2}}$. $\bf \color{red} (1)$ is derived from the fact that $x-x^2\leq \frac{1}{4} \ \forall x\in \mathbb{R}$

Let $a,b,c>0$ such that $a \geq bc^2$ , $b \geq ca^2$ and $c \geq ab^2$ . $(1)$ Find the maximum value of $(a-bc^2)(b-ca^2)(c-ab^2)$. $(2)$ Find the maximum value of $a^2b^2c^2(a-bc^2)(b-ca^2)(c-ab^2)$. here

Let $a,b,c>0$ such that $a \geq bc^2$ , $b \geq ca^2$ and $c \geq ab^2$ . Prove that $(1)$ $(a-bc^2)(b-ca^2)(c-ab^2)\leq\frac{\sqrt{3}}{243}$. $(2)$ $a^2b^2c^2(a-bc^2)(b-ca^2)(c-ab^2)\leq\frac{648\sqrt{15}}{390625}$.

ComplexPhi wrote: Let $a,b,c>0$ such that $a \geq bc^2$ , $b \geq ca^2$ and $c \geq ab^2$ . Find the maximum value that the expression : $$E=abc(a-bc^2)(b-ca^2)(c-ab^2)$$can acheive. Let $ a,b,c$ be positive real numbers such that $a>b^2,b>c^2 .$ Find the maximum value of $(a-b^2)(b-c^2)(c-a^2) .$