For $t\geq0,x\geq 0$, I want to prove the following inequality
$$\int_0^t s(t-s)(1+s)^{x-1}(1+t-s)^{x-1}(2t-3s)\log(1+s)\,ds\geq 0.$$
I have verified the above inequality in Mathematica by numerical integration. But I don't know how to prove it. One idea is that maybe the Hermite-Hadamard-Fejer inequality could be used.
Hermite-Hadamard-Fejer:
$f:[a,b]\to \mathbb{R}$ is a concave function, $w:[a,b]\to \mathbb{R} $ is a non-negative, integrable function and symmetric about $\frac{a+b}{2}$, then the inequality holds
$$\frac{f(a)+f(b)} 2 \int_a^b w(s) \, ds \leq\int_a^b f(s)w(s) \, ds\leq f\left(\frac{a+b} 2 \right) \int_a^b w(s)\,ds.$$
If we let $f(s)=(2t-3s)\log(1+s)$, $w(s)=s(t-s)(1+s)^{x-1}(1+t-s)^{x-1}$. However, by above inequality, I can only get a negative lower bound...