What's maximum value of $x (1-x^2)$ for $0 < x <1$?

Let $x=\frac{a}{\sqrt3}.$

Thus, by AM-GM $$x(1-x^2)=\frac{1}{3\sqrt3}(3a-a^3)=\frac{1}{3\sqrt3}(2-(a^3+2-3a))\leq$$ $$\leq\frac{1}{3\sqrt3}(2-(3\sqrt[3]{a^3\cdot1^2}-3a))=\frac{2}{3\sqrt3}.$$ The equality occurs for $a=1$ or $x=\frac{1}{\sqrt3},$ which says that we got a maximal value.

We can use AM-GM also by the following way. $$x(1-x^2)=\sqrt{\frac{1}{2}\cdot2x^2(1-x^2)^2}\leq$$ $$\leq\sqrt{\frac{1}{2}\left(\frac{2x^2+1-x^2+1-x^2}{3}\right)^3}=\frac{2}{3\sqrt3}.$$

Notice the A.M/G.M theorem says:

$a_1 + a_2 + .. + a_n \ge n\sqrt[n]{a_1a_2...a_n}$. And that $a_1 + a_2 + .. + a_n = n\sqrt[n]{a_1a_2...a_n}$ if and ONLY IF $a_1 = a_2 = .... = a_n$.

Your calculations for the inequalities were correct. But you ignored the requirements for equality.

So by your calculations.

$v \le \frac 9{16}$ ... which is true.

And $v = \frac 9{16}$ if and only if $x^2 = 1 -x^2 = 1+x = 1-x$. .... which never happens. As that is never possible we know that $v$ doesn't ever equal $\frac 9{16}$ . But we do know that $v < \frac 9{16}$ always. So this is an upper bound of $v$.

But it need not be a maximum. If a maximum does exist it is less than $\frac 9{16}$.

That's .... all true.

Also by your calculations.

$v \le \frac {2}{27^{\frac 12}} = \frac 2{3\sqrt 3} < \frac 9{16}$.

And $v = \frac 2{3\sqrt 3}$ if and only if $2x^2 = 1-x^2 =1-x^2$ or if and only if $x =\pm \frac 1{\sqrt 3}$.

As that is possible if $x = \frac 1{\sqrt 3}$. So we have $v \le \frac 2{3\sqrt 3}$ with equality holding if and only if $x = \frac 1{\sqrt 3}$.

So that is the maximum value.

$v \le \frac 2{3\sqrt 3} < \frac 9{16}$.