Prove that $\int_{0}^{1}{f^{2}(x)dx}\leq \frac{4}{3}\left(\int_{0}^{1}{f(x)dx}\right)^2$ Let $f(x)$ be a concave nonnegative function on $[0,1]$
Prove that 
$$\displaystyle \int\limits_{0}^{1}{f^{2}(x)dx}\leq \frac{4}{3}\left(\int\limits_{0}^{1}{f(x)dx}\right)^2$$
My friend tian_275461 told me we even have the general result
Let $f(x)$ be a concave nonnegative function on $[a,b]$,If $p>1$
$$ \frac{2^{p}}{p+1}\left(\frac{1}{b-a}\int\limits_{a}^{b}{f(x)dx}\right)^{p}\geq \frac{1}{b-a}\int\limits_{a}^{b}{f^{p}(x)dx} $$ 
If $0<p<1 $,the reverse inequality holds.
I don't know how to deal with such function which is concave. 
 A: Well,my friend found a proof in case $ a=0,b=1 $,I think this method can be used in this original inequality.
Here his the solution.
With out loss of generally,we consider the case $f(0)=f(1)=0$,and $f(x)$ has order continuous derivative,Therefore
$$f''(x)\leq 0 $$
Thus
$$ f(x)=-\int_{0}^{1}{K(x,t)f''(t)dt} $$
Where $K(x,t)$ is Green function.
$$ K(x,t)=\left\{
  \begin{array}{ll}
   t(1-x)  & \hbox{$0\leq t\le x\le 1$} \\
    x(1-t)  & \hbox{$0\leq x\le t\le 1$}
  \end{array}
\right.
$$
Then by Minkowski inequality,we have
\begin{align}
\left(\int_{0}^{1}{f^{p}(x)dx}\right)^{\frac{1}{p}}&=\left(\int_{0}^{1}{\left(\int_{0}^{1}{K(x,t)(-f''(t))dt}\right)^{p}dx     }\right)^{\frac{1}{p}}\\
&\leq \int_{0}^{1}{\left(\int_{0}^{1}{K^{p}(x,t)(-f''(t))^{p}dx}\right)dt}\\
&=\frac{1}{(p+1)^{\frac{1}{p}}}\int_{0}^{1}{t(1-t)|f''(t)|dt}
\end{align}
On the other hand
\begin{align}
\int_{0}^{1}{f(x)dx}&=-\int_{0}^{1}{\int_{0}^{1}{K(x,t)f''(t)dt}  dx}\\
&=-\int_{0}^{1}{\int_{0}^{1}{K(x,t)f''(t)dx}  dt}\\
&=-\frac{1}{2}\int_{0}^{1}{t(1-t)f''(t)dt}  
\end{align}
Therefore
$$ \int_{0}^{1}{f^{p}(x)dx}\leq \frac{2^p}{p+1}\left(\int_{0}^{1}{f(x)dx}\right)^p $$ 
