# An integral inequality

1. Jul 13, 2013

### 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.