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.


$f:[a,b]\to \mathbb{R}$ 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...

  • 1
    $\begingroup$ First things first, I would enforce the substitution $s=tu$ in order to deal with $$ J(x,t)=\int_{0}^{1}(1-u)^{x-1}(1+t(1-u))^{x-1}\color{red}{(2u-3u^2)}\log(1+tu)\,du$$ then apply integration by parts in order to convert the red term into a nicer $u^2(1-u)$. $\endgroup$ – Jack D'Aurizio Dec 10 '17 at 17:34
  • $\begingroup$ @JackD'Aurizio Good idea. However, I dont expect the function $J(x,t)$ have the explicit expression. $\endgroup$ – Hanlin Dec 10 '17 at 17:49

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