# An integral inequality

#### amirmath

I want to know that is it possible to show that
$$\int_{0}^{T}\Bigr(a(t )\Bigr)^{\frac{p+1}{2p}}dt\leq C\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}}$$
for some $C>0$ where $a(t)>0$ and integrable on $(0,T)$ and $p\in(\frac{1}{2},1)$. It is worth noting that this range for $p$ yields $\frac{p+1}{2p}>1$. In the case $p>1$ we have $\frac{p+1}{2p}<1$ and the Holder's inequality can be applied to obtain the result immediately. But in the first case I don't know the validity of the inequality.

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#### fresh_42

Mentor
2018 Award
The crucial point is whether $C$ may depend on $T$ or not. E.g. $p=\frac{2}{3}\, , \,a(t)=t$.

"An integral inequality"

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