# Proof for this integral inequality

I am trying to prove that for $$p>1$$, $$f(x)$$ be non-negative and non-increasing function, then

$$, \left( \frac{1}{x}\int_{0}^{x}f(t)F^{p-1}(t)dt\right) -\frac{p-1}{p}\left( \frac{1}{x}\int_{0}^{x}F^{p}(t)dt\right) \leq \frac{1}{p}F^{p}(x) \tag 1$$ where $$F(x)=\frac{1}{x}\int_{0}^{x}f(t)dt\text{.}$$

My proof started as follows

Since $$f(t)$$ is a decreasing function, then $$F^{p}(x)$$ is also a decreasing operator as follows \begin{align*} \left( F^{p}(x)\right) ^{\prime } &=\left( \left( \frac{1}{x} \int_{0}^{x}f(t)dt\right) ^{p}\right) ^{\prime }\\ &=pF^{p-1}(x)F^{\prime }(x) \\ &=pF^{p-1}(x)\left[ \frac{1}{x}f(x)-\frac{1}{x^{2}}\int_{0}^{x}f(t)dt\right] , \end{align*} but since $$f(t)$$ is a decreasing, then $$\begin{equation*} \frac{1}{x^{2}}\int_{0}^{x}f(t)dt\geq \frac{1}{x^{2}}xf(x)=\frac{1}{x}f(x), \end{equation*}$$ substituting this, leads to $$\begin{eqnarray*} \left( F^{p}(x)\right) ^{\prime } &\leq &pF^{p-1}(x)\left[ \frac{1}{x}f(x)- \frac{1}{x}f(x)\right] \\ &=&0, \end{eqnarray*}$$ then, we can write that $$\begin{eqnarray*} \frac{1}{x}\int_{0}^{x}f(t)F^{p-1}(t)dt &\geq &\left( \frac{1}{x} \int_{0}^{x}f(t)dt\right) F^{p-1}(x) \\ &=&F(x)F^{p-1}(x)=F^{p}(x) \end{eqnarray*}$$

and I got stuck after that, Any suggestion to complete the proof?

I'll re-label $$F(x)$$ into $$M(x)$$ not to be confused with anti-derivative of $$f$$. $$\left( \frac{1}{x}\int_{0}^{x}f(t)M^{p-1}(t)\,\mathrm{d}t\right) -\frac{p-1}{p}\left( \frac{1}{x}\int_{0}^{x}M^{p}(t)\,\mathrm{d}t\right) \leq \frac{1}{p}M^{p}(x)$$ $$\hbox{ where }M(x)=\frac{1}{x}F(x) \hbox{ and }F(x)=\int_{0}^{x}f(t)\,\mathrm{d}t,$$ so we need to prove $$\left( \frac{1}{x}\int_{0}^{x} \frac{1}{t^{p-1}}f(t)F^{p-1}(t)\,\mathrm{d}t\right) -\frac{p-1}{p}\left( \frac{1}{x}\int_{0}^{x}\frac{1}{t^p}F^{p}(t)\,\mathrm{d}t\right) \leq \frac{1}{px^p}F^{p}(x)$$ Let $$x>0$$ first and $$p_1=p-1>0$$ $$\left( (p_1+1)\int_{0}^{x} \frac{1}{t^{p_1}}f(t)F^{p_1}(t)\,\mathrm{d}t\right) -p_1\left( \int_{0}^{x}\frac{1}{t^{p_1+1}}F^{p_1+1}(t)\,\mathrm{d}t\right) \leq \frac{1}{x^{p_1}}F^{p_1+1}(x)$$ Taking $$-p_1\left( \int\limits_{0}^{x}\frac{1}{t^{p_1+1}}F^{p_1+1}(t)\,\mathrm{d}t\right)$$ by parts $$-p_1\left( \int_{0}^{x}\frac{1}{t^{p_1+1}}F^{p_1+1}(t)\,\mathrm{d}t\right)= \int_{0}^{x}F^{p_1+1}(t)\,\mathrm{d}\left(\frac{1}{t^{p_1}}\right) \\=\left[\frac{1}{t^{p_1}}F^{p_1+1}(t)\right]_{t=0}^x- (p_1+1)\int\limits_0^x\frac{1}{t^{p_1}}F^{p_1}(t)f(t)\,\mathrm{d}t$$ almost everything cancels, and we only need to prove $$-\lim\limits_{t\to +0}\frac{1}{t^{p_1}}F^{p_1+1}(t)\le 0$$ which is obvious since everything within the limit sign is positive.
However, I'm not sure how we are to prove $$-\lim\limits_{t\to -0}\frac{1}{t^{p_1}}F^{p_1+1}(t)\ge 0$$ for the case $$x<0$$.
• Actually, we do not need it for negative $x$. Jun 23, 2020 at 9:58