I was reading the solution to a functional inequality in an article when the author made the following remarks without giving any proof: let $f(x): [0, \infty]\to[0, \infty]$ be such that $$\left(\int_0^t f(x)dx\right)^2 \ge \int_0^t f(x)^3dx$$ for all $t>0$. Then, the following two statements are true:
- $\int_0^t f(x)dx \le \frac{t^2}{2}$;
- $\int_0^t f(x)^\gamma dx \le \frac{1}{\gamma +1}\left(2\int_0^t f(x)dx\right)^{(\gamma + 1)/2}$ for all positive $t$ and $\gamma \in [1,3]$.
Again, there is no proof in the article, so I don't know if these statements are fairly easy to prove or are very involved. One thing that might be worth mentioning is that all these inequalities become exact when $f(x)=x$. I am wondering if anyone has an idea or have seen these before.
While I do not have the answer to the question, one can use Hölder's inequality to get some weaker bounds :
$$ \int_0^t f(x).1 \,\mathrm{d}x \leq \left(\int_0^t f(x)^p \,\mathrm{d}x\right)^{1/p}\left(\int_0^t 1^q\,\mathrm{d}x\right)^{1/q} $$
with parameters $p=3$, $q=3/2$ gives, after a few manipulations
$$ \int_0^t f(x) \,\mathrm{d}x \leq t^2, $$
which differs from the first statement by a factor $2$.
$$ \int_0^t f(x)^a.f(x)^b \,\mathrm{d}x \leq \left(\int_0^t f(x)^{ap} \,\mathrm{d}x\right)^{1/p}\left(\int_0^t f(x)^{bq}\,\mathrm{d}x\right)^{1/q} $$
with $a, b>0$, setting $ap=3$, $bq=1$ and $a+b=\gamma$, gives a unique choice of parameters (namely, $a=(3\gamma-3)/2$ and $b=(3-\gamma)/2$). One then obtains
$$ \int_0^t f(x)^\gamma \,\mathrm{d}x \leq \left(\int_0^t f(x) \,\mathrm{d}x\right)^{(\gamma+1)/2}, $$
which again is one constant away from the second statement.