As you wrote in a comment we have the Hölder inequality:
If $h\in L^p(\mathbb{R})$ and $k\in L^{p'}(\mathbb{R})$, where $1/p+1/p'=1$ and $1\leq p,\,p'\leq +\infty$ then
$$
\int_\mathbb{R} |h(x)k(x)|dx\leq \left(\int_\mathbb{R} |h(x)|^pdx\right)^{1/p}\left(\int_\mathbb{R} |k(x)|^{p'}dx\right)^{1/p'}
$$
In our case we look at $f\in L^1(\mathbb{R})$ and $g\in L^{p}(\mathbb{R})$, and we wish to estimate their convolution (in a point-wise sense)
$$f*g(t) =\int f(x) g(t-x)\,dx.$$
However, since $1/p+1/1\ne1$ something has to be done in order to achieve a proper estimate.
One thing we always can do is to find the exponent $p'$ appearing in Hölder's inequality that correspond to $p$ (the so called conjugate exponent of $p$): it is the number $p'=p/(p-1)$. With this choice we have
$$
|f(x)|= |f(x)|^1= |f(x)|^{1/p+ 1/p'}= |f(x)|^{1/p}\cdot|f(x)|^{1/p'}
$$
which leads to
\begin{eqnarray*}
\int|f(x)||g(t-x)|dx &=& \int |f(x)|^{1/p'}\cdot|f(x)|^{1/p}|g(t-x)|dx\\ &\leq&\left(\int |f(x)|^{p'/p'}\right)^{1/p'}\left(\int |f(x)|^{p/p}|g(t-x)|^p\,dx\right)^{1/p}\\
&=&\|f\|_{L^1}^{1/p'}\left(\int |f(x)||g(t-x)|^p\,dx\right)^{1/p}
\end{eqnarray*}
by Hölder's inequality - this is bounded because both $f$ and $|g|^p$ belongs to $L^1$, and $L^1$ is closed under convolution.
For your last query regarding validity of $\left(\int |f(x)| dx\right)^{1/p} \leq \int |f(x)|^{1/p}$ - This is false.
It is true for the counting measure on $\mathbb{Z}$, but not on $\mathbb{R}$.
Counter example:
Define $f$ as piece-wise constant and unbounded as follow. For each integer $n
>0$ we write $f(x) = n$ if $x\in(n,n+1/n^2)$, elsewhere we set $f(x)=0$, then
$$\int f(x) dx = \sum_{n>0} \int_n^{n+1/n^2}n\, dx = \sum_{n>0} n\cdot\frac{1}{n^2} = \infty$$
while
$$\int f(x) dx = \sum_{n>0} \int_n^{n+1/n^2}n\, dx = \sum_{n>0} n\cdot\frac{1}{n^2} = \infty$$
while
$$\int |f(x)|^{1/p} dx =\sum_{n>0} \int_n^{n+1/n^2}n^{1/p}\, dx = \sum_{n>0} n^{1/p}\cdot\frac{1}{n^2} < \infty$$
because $2-1/p>1$ for $p>1$.
