An apparently simple but somehow unexpected inequality between integrals Let $f:[0,1]\to[0,1]$ be any Lebesgue integrable function. Then
$$I_1= \left(\int_0^1f(x)\sqrt x dx\right)^2 \leq \frac{4}{3}\int_0^1f(x)x^2dx=I_2\,.\tag{$\star$}$$
At first sight, I would have expected the reverse inequality, as $x^2\leq \sqrt x$ on $[0,1]$ and I didn't think that taking the square of the LHS would have been enough to compensate. However, it is easy to see that if $f=\chi_{[0, b]}$, for some $b\in(0, 1]$, where $\chi_A$ is the characteristic function of the set $A$, then
$I_1=I_2 = \frac{4}{9} b^3$. On the other hand, if we let $[a, b]\subseteq[0,1]$, then taking $f=\chi_{[a, b]}$ we have
$I_1 = \frac{4}{9}(b^{3/2}-a^{3/2})^2$ and $I_2 = \frac{4}{9}(b^{3}-a^{3})$. It is not hard to prove (for instance using the sub-additivity of $\sqrt\cdot$) that $I_1\leq I_2$. Then for $\epsilon\in(0,1]$ and $f=\epsilon\chi_{[a,b]}$ it is clear that the same result holds true, and actually $I_1/I_2$ will scale as $\epsilon$, so it is maximised for $\epsilon=1$. (Note that allowing $\epsilon>1$ the inequality would not be true, but we would not be in the hypothesis $f\subseteq [0,1]$.) So, for all $f$ which is a simple function (non-zero and constant on finitely many intervals) we get that ($\star$) is satisfied. Taking the limit of simple functions we can obtain any measurable function, and so conclude. [This is wrong! See edit.]
Now, as the inequality looks quite simple, I would expect that there is a much simpler proof for it, for instance something making use of Hölder's or Jensen's inequality. I might have missed something trivial, but so far I didn't manage to do it, and I have the feeling that I am not really grasping what is going on.
Notice that we can restate ($\star$) as $E_1^2\leq E_2$, where
$$E_1 = \frac{3}{2}\int_0^1 f(x)\sqrt x dx$$
and $$E_2 = 3 \int_0^1 f(x) x^2 dx\,.$$
The reason for this transform is that then you can see $\frac{3}{2}\sqrt x$ and $3 x^2$ as probability densities on $[0,1]$, and so rewrite everything in terms of expectations of $f$. However, so far I couldn't find any smart trick for a few-lines proof of ($\star$), without having to pass via a sequence of simple functions converging to $f$.
Edit. I've actually just realised that the proof I had provided does not work, as if $f=\sum_i f_i =  \chi_{(a_i, b_i)}$, then we just have
$$I_1 = \left(\sum_i\int_0^1 f_i(x)\sqrt xdx\right)^2 \leq 2\sum_i \left(\int_0^1 f_i(x)\sqrt xdx\right)\leq 2 I_2\,.$$
However the proof from @Sangchul Lee shows that the inequality $I_1\leq I_2$ holds.
 A: By substituting $x = u^{2/3}$ and letting $g(u) = f(u^{2/3})$, we get
$$ E_1 = \int_{0}^{1} g(u) \, \mathrm{d}u \qquad\text{and}\qquad E_2 = \int_{0}^{1} 2u g(u) \, \mathrm{d}u. $$
From this,
\begin{align*}
E_1^2
= \left( \int_{0}^{1} g(u) \, \mathrm{d}u\right)^2
= \int_{0}^{1}\int_{0}^{1} g(u)g(v) \, \mathrm{d}u\mathrm{d}v
\leq \int_{0}^{1}\int_{0}^{1} g(\max\{u, v\}) \, \mathrm{d}u\mathrm{d}v.
\end{align*}
Indeed, the last line follows by writing $g(u)g(v) = g(\max\{u,v\})g(\min\{u,v\})$ and then noting that $0 \leq g \leq 1$. Finally, by substituting $r = \max\{u, v\}$, the last integral reduces to
\begin{align*}
\int_{0}^{1} 2r g(r) \, \mathrm{d}r = E_2.
\end{align*}
A: Here is another answer along the lines of that of Sangchui Lee.
By Funini's theorem and the fact that $0\leq f\leq 1$,
\begin{align}
\int^1_0\int^1_0f(x)f(y)\sqrt{xy}\,dydx&=\int^1_0\int^1_0f(x\wedge y)f(x\vee y)\sqrt{x\wedge y}\sqrt{x\vee y}\,dydx\\
&\leq\int^1_0\int^1_0f(x\vee y)\sqrt{x\wedge y}\sqrt{x\vee y}\,dydx\\
&=\int^1_0\Big(\int^x_0 +\int^1_x\Big)f(x\vee y)\sqrt{x\wedge y}\sqrt{x\vee y}\,dydx\\
&=\int^1_0\int^x_0f(x)\sqrt{x}\sqrt{y}\,dy\,dx+\int^1_0\int^1_xf(y)
\sqrt{x}\sqrt{y}\,dydx\\
&=\int^1_0f(x)\sqrt{x}\Big(\int^x_0\sqrt{y}\,dy\Big)\,dx+\int^1_0f(y)\sqrt{y}\Big(\int^y_0\sqrt{x}\,dx\Big)\,dy\\
&= \frac43\int^1_0f(u)u^2\,du
\end{align}
