Let $p\in (1,+\infty)$ and $q$ its conugate exponent, that is, the real number $q$ such that $\frac 1p+\frac 1q=1$. Let $f\colon (0,+\infty)\to\mathbb R\in L^p(0,+\infty)$ and
$F(x)=\frac 1x\int_{\left]0,x\right[}f(t)dt$. Then $F\in L^p(0,1)$ and $\lVert F\rVert_{L^p}\leq q\lVert f\rVert f_{L^p}$.
Pick $0<\alpha<\frac 1q$, we will specifie it later.
We start using Hölder inequality
\begin{align*}
|xF(x)|&=\left|\int_{\left]0,x\right[}f(t)t^{\alpha}t^{-\alpha}dt\right|\\\
&\leq \left(\int_{\left]0,x\right[}\left| f(t) t^{\alpha}\right|^pdt\right)^{\frac 1p}
\left(\int_{\left]0,x\right[}t^{-q\alpha}\right)^{\frac 1q}\\\
&=\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt\right)^{\frac 1p}
\left(\frac 1{1-q\alpha}x^{1-q\alpha}\right)^{\frac 1q}\\\
&=(1-q\alpha)^{-\frac 1q}\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{\frac 1{q}}dt\right)^{\frac 1p}x^{\frac 1q-\alpha},
\end{align*}
hence
\begin{align*}
|F(x)|&\leq (1-q\alpha)^{-\frac 1q}\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt\right)^{\frac 1p}x^{\frac 1q-\alpha -1}\\\
&=(1-q\alpha)^{-\frac 1q}\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt\right)^{\frac 1p}x^{-\alpha -\frac 1p}
\end{align*}
and
$$|F(x)|^p\leq (1-q\alpha)^{-\frac pq}\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt x^{-p\alpha-1}.$$
Integrating and using Fubini's theorem, we get
\begin{align*}
\int_{(0,+\infty)}|F(x)|^pdx&\leq (1-q\alpha)^{-\frac pq}\int_{(0,\infty)}\int_{(0,x)}
\left| f(t)\right|^p t^{p\alpha}x^{-p\alpha-1}dtdx\\\
&=(1-q\alpha)^{-\frac pq}\int_{(0,+\infty)}\int_{(t,+\infty)}\left| f(t)\right|^p t^{p\alpha}x^{-p\alpha-1}dxdt\\\
&=(1-q\alpha)^{-\frac pq}\int_{(0,+\infty)}\left| f(t)\right|^p t^{p\alpha}\left(\int_{(t,+\infty)}x^{-p\alpha-1}dx\right)dt,
\end{align*}
thus
$$\int_{(0,+\infty)}|F(x)|^pdx \leq(1-q\alpha)^{-\frac pq}\int_{(0,+\infty)}\left|f(t)\right|^p t^{p\alpha}\frac 1{p\alpha}t^{-p\alpha}dt$$
and finally
$$\int_{(0,+\infty)}|F(x)|^pdx\leq (1-q\alpha)^{-\frac pq}(p\alpha)^{-1}\int_{(0,+\infty)}\left| f(t)\right|^p dt.$$
Now, we pick $\alpha:=\frac 1{pq}<\frac 1q$, to get
$$(1-q\alpha)^{-\frac pq}(p\alpha)^{-1} =(1-\frac 1p)^{-\frac 1q}q=
q^{\frac pq}q=q^{1+p\left(1-\frac 1p\right)}=q^p,$$
which gives $\lVert F\rVert_{L^p}\leq q\lVert f\rVert_{L^p}$.
This question deals with the case of the best constant.