Context: Proving integral inequalities about posterior distributions following different sequences of binary signals. The proofs come down to the following inequalities.
Let $\psi(x)$ be a concave probability density function of a random variable whose support is a connected subset of [0,1].
Can one prove that for all $a$ such that $\Psi(a) > 0$ (with $\Psi(x)$ being the cumulative density function),
$$ \int_0^a x^2\psi(x)\,dx \int_0^a x^2\psi(x)\,dx \leqslant \int_0^a x^3\psi(x)\,dx \int_0^a x\psi(x)\,dx $$ and $$ \int_0^a x\psi(x)\,dx \int_0^a x(1-x)\psi(x)\,dx \leqslant \int_0^a \psi(x)\,dx \int_0^a x^2(1-x)\psi(x)\,dx $$?
I suspect a more general result could be proven, but for my purpose these cases suffice.
For the first inequality, the concavity is not needed.
Also the condition $\Psi(a)>0$ is irrelevant, since the inequality reads $0\le 0$ otherwise.
Yet let us prove it for $\Psi(a)>0$. Define the probability density on $[0,a]$: $f(x) = x\psi(x)/K$, where $K = \int_0^a x\psi(x)dx$. For a random variable $\xi$ with this density, $\mathsf E[\xi^2]\ge \left(\mathsf E[\xi]\right)^2$, which is equivalent to the first inequality.
The second inequality seems false for $\psi(x) = 2\mathbf{1}_{[1/2,1]}$, $a=1$, reading $3/16\le 1/8$. Correct me if I'm wrong.
More on the second inequality. Let, for definiteness, $a=1$. Then it is equivalent to $$ \frac{\int_0^1 x\psi(x) dx}{\int_0^1 \psi(x) dx}\le \frac{\int_0^1 x\psi_1(x) dx}{\int_0^1 \psi_1(x) dx} $$ with $\psi_1(x) = x(1-x)\psi(x)$. So if it were true, it would mean that multiplying $\psi$ by $x(1-x)$ increases the expectation of corresponding random variable. But $x(1-x)$ is symmetric. So changing $\psi(x)$ to $\psi(1-x)$, we arrive at a contradiction.