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
$$2x^2+(1-x^2)+(1-x^2)
>3 (2v^2)^{\frac{1}{3}}
\implies \frac{2}{(27)^{\frac{1}{2}}} > v.
$$

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$(1)$
