Functional inequalities involving the condition $\left(\int_0^t f(x)dx\right)^2 \ge \int_0^t f(x)^3dx$ 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 $t&gt;0$. 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 :


Hölder's inequality


$$
\int_0^t f(x).1 \,\mathrm{d}x \leq \left(\int_0^t f(x)^p \,\mathrm{d}x\right)^{1/p}\left(\int_0^t 1^q\,\mathrm{d}x\right)^{1/q}
$$

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 $2$.


Likewise, applying Hölder's inequality


$$
\int_0^t f(x)^a.f(x)^b \,\mathrm{d}x \leq \left(\int_0^t f(x)^{ap} \,\mathrm{d}x\right)^{1/p}\left(\int_0^t f(x)^{bq}\,\mathrm{d}x\right)^{1/q}
$$

with $a, b&gt;0$, 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.
While I do not have the answer to the question, one can use Hölder's inequality to get some weaker bounds :


Hölder's inequality


$$
\int_0^t f(x).1 \,\mathrm{d}x \leq \left(\int_0^t f(x)^p \,\mathrm{d}x\right)^{1/p}\left(\int_0^t 1^q\,\mathrm{d}x\right)^{1/q}
$$

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 $2$.


Likewise, applying Hölder's inequality


$$
\int_0^t f(x)^a.f(x)^b \,\mathrm{d}x \leq \left(\int_0^t f(x)^{ap} \,\mathrm{d}x\right)^{1/p}\left(\int_0^t f(x)^{bq}\,\mathrm{d}x\right)^{1/q}
$$

with $a, b&gt;0$, 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.