<s>[INST] Maximize $\int_{0}^{1} f(x)^5 dx$ over all $f\colon[0,1]\to[-1,1]$ with $\int_{0}^{1} f(x)^3 dx=\int_{0}^{1} f(x) dx= 0$. 
Maximize $\int_{0}^{1} f(x)^5 dx$ over all $f\colon[0,1]\to[-1,1]$ with $\int_{0}^{1} f(x)^3 dx=\int_{0}^{1} f(x) dx= 0$.

I'm not even sure where to start with this problem. Any hints would be appreciated.
 [/INST] The maximum value of $\int_0^1 f(x)^5 \mathrm{d}x$ is $\frac{1}{16}$.
We first prove that $\frac{1}{16}$ is an upper bound.
Note that if $t\le 1$ then $t^5\le \frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}$. Indeed, 
$$t^5 - \left(\frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}\right) = (t-1)\left(t^2+\frac t2 - \frac 14 \right)^2 \le 0.$$
Putting $t=f(x)$ and integrating we obtain
$$\int_0^1 f(x)^5 \mathrm{d}x \le \int_0^1 \left(\frac 54 f(x)^3 - \frac{5}{16}f(x)+\frac{1}{16} \right) \mathrm{d}x = \frac{1}{16},$$
since by assumption $\int_0^1 f(x)\mathrm{d}x =\int_0^1 f(x)^3\mathrm{d}x=0$.
The following example shows that the bound $\frac{1}{16}$ is optimal:
$$f(x)=\begin{cases} 
\frac 14 (-1-\sqrt 5) & \text{ for } 0\le x < \frac 25 \\
\frac 14(-1+\sqrt 5) & \text{ for } \frac 25 \le x < \frac 45 \\
1 & \text{ for } \frac 45 \le x \le 1 \\
\end{cases}$$
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