Show that if $\int_0^1f(x)dx=a$, then $\int_0^1\sqrt{f(x)}dx\ge a^{2/3}$ 
$f$ is continuous on $[0,1]$ and there is $a>0$ such that, $0\le f(x)\le a^{2/3}$ for $x\in[0,1]$. Show that if $\displaystyle\int_0^1f(x)dx=a$, then $\displaystyle\int_0^1\sqrt{f(x)}dx\ge a^{2/3}$

To get an expression like $\int\sqrt{f(x)}dx$, I take $f^{\frac{1}{2p}}$ instead of $f$, since I apply Hölder-Inequality.(and if $\frac{1}{p}+\frac{1}{q}=1$, then $q=\frac{p}{p-1}$)
$a=\displaystyle\int_0^1 f=\displaystyle\int_0^1f^{\frac{1}{2p}}f^{1-\frac{1}{2p}}\le\Big(\displaystyle\int_0^1\sqrt{f(x)}\Big)^{\frac{1}{p}}\Big(\displaystyle\int_0^1(f^{\frac{2p-1}{2p}})^{\frac{p}{p-1}}\Big)^{\frac{p-1}{p}}$
Hence $a^p\le\Big(\displaystyle\int_0^1\sqrt{f(x)}\Big)\Big(\displaystyle\int_0^1f^{\frac{2p-1}{2p-2}}\Big)^{p-1}$
but I can't write the second integral in terms of $a$, do you have an idea ?
Thanks in advance.
 A: From
$$a^p\leqslant\left(\displaystyle\int_0^1\sqrt{f(x)}\right)\left(\displaystyle\int_0^1f^{\large\frac{2p-1}{2p-2}}\right)^{p-1},$$
it is only a short step. We use $0 \leqslant f \leqslant a^{2/3}$ to further obtain
$$\begin{align}
a^p &\leqslant\left(\displaystyle\int_0^1\sqrt{f(x)}\right)\left(\displaystyle\int_0^1f^{\large\frac{2p-1}{2p-2}}\right)^{p-1}\\
&\leqslant \left(\int_0^1 \sqrt{f(x)}\right)\left(a^{\large\frac{2}{3}\cdot\frac{2p-1}{2p-2}}\right)^{p-1}\\
&= \left(\int_0^1 \sqrt{f(x)}\right)\cdot a^{\large\frac{2p-1}{3}},
\end{align}$$
from which rearranging yields
$$a^{\large\frac{p+1}{3}} \leqslant \int_0^1 \sqrt{f(x)}.$$
Now letting $p \downarrow 1$ yields the result.
A: $$a=\Vert \sqrt{f}\Vert^2_{L_2[0,1]}\\
\qquad\qquad\qquad \leq \Vert \sqrt{f}\Vert_{L_1[0,1]}\Vert \sqrt{f}\Vert_{L_\infty[0,1]}\\
\\\qquad \leq a^\frac{1}{3}\Vert \sqrt{f}\Vert_{L_1[0,1]}$$
Therefore, $$\Vert\sqrt{f}\Vert_{L_1[0,1]}\geq a^\frac{2}{3}.$$
A: I will present $2$ ways:
$\textbf{1st Way:}$
$$\int_0^1f=a\implies\int_0^1 \frac{f}{a^{1/3}}=a^{2/3}$$
the $2$nd integral can be written as:
$$\int_0^1\sqrt\frac{f}{a^{2/3}}\cdot\sqrt{f}=a^{2/3}$$
but the first term inside the integral is $\le1$
Hence $$\int_0^1\sqrt{f}\ge a^{2/3}$$
$\textbf{2nd Way:}$
Basically it is the same as Daniel Fischer’s answer but I don’t fully replace $f$ by $a^{2/3}$, I just split it up in $\sqrt{f}$ term and subsitute the remaining term
$$\begin{align}
a^p &\leqslant\left(\displaystyle\int_0^1\sqrt{f(x)}\right)\left(\displaystyle\int_0^1f^{\large\frac{2p-1}{2p-2}}\right)^{p-1}\\
&\leqslant \left(\int_0^1 \sqrt{f(x)}\right)\left(\sqrt{f(x)}\cdot f^{\frac{p}{2p-2}}\right)^{p-1}\\
&\leqslant \left(\int_0^1 \sqrt{f(x)}\right)\left(\sqrt{f(x)}\cdot a^{\frac23\cdot\frac{p}{2p-2}}\right)^{p-1}\\
&=\left(\int_0^1 \sqrt{f(x)}\right)\left(\int_0^1 \sqrt{f(x)}\right)^{p-1}\cdot a^{\frac23\cdot\frac{p}{2p-2}\cdot(p-1)}
\end{align}$$
this means we have;
$$\iff a^p\le\left(\int_0^1 \sqrt{f(x)}\right)^p\cdot a^{p/3}$$
$$\iff a^{\frac{2p}{3p}}=a^{\frac23}\le\left(\int_0^1 \sqrt{f(x)}\right)$$
