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?
 A: 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 $x<0$.
