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.

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 $$ \int_{0}^{1}{f^{p}(x)dx}\leq \frac{2^p}{p+1}\left(\int_{0}^{1}{f(x)dx}\right)^p $$