What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtain
\begin{equation}
x^2+(1-x^2)+(1+x)+(1-x)
>4 (v^2)^{\frac{1}{4}}
\implies \frac{9}{16} \ge v.
\end{equation}
Therefore the max value is $\frac{9}{16}$.
Method 2
$$
v=x (1-x^2)
\implies 2v^2= 2x^2 (1-x^2)^2
$$
With the AM-GM-inequality
$$2x^2+(1-x^2)+(1-x^2)
>3 (2v^2)^{\frac{1}{3}}
\implies \frac{2}{(27)^{\frac{1}{2}}} > v.
$$
 A: 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.
$$x(1-x^2)=\sqrt{\frac{1}{2}\cdot2x^2(1-x^2)^2}\leq$$
$$\leq\sqrt{\frac{1}{2}\left(\frac{2x^2+1-x^2+1-x^2}{3}\right)^3}=\frac{2}{3\sqrt3}.$$
A: 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.
$v \le \frac 2{3\sqrt 3} < \frac 9{16}$.
A: Hint:
$$\dfrac{ax+b(1+x)+c(1-x)}3\ge ?$$
for $a,b,c>0$
Set $a+b-c=0\ \ \  \ (1)$
The equality will occur if $ax=b(x+1)=c(x-1)=k$(say)
Put the values of $a,b,c$ in$(1)$
