Minkowski's Inequality I am wondering how to prove inequality 
$$\left\| \int f(.,y)dy \right\|_{p}\leq \int \left\| f(.,y) \right\|_{p}dy~~~~?$$ 
Here, $f$ is an integrable function on $\mathbb{R}^n$  and $\displaystyle \left\|f\right\|_{p}=\left( \int|f|^pdx \right)^{1/p}$ for $p\geq 1.$
 A: Let's start with this version of Minkowski's inequality:
$$
\|f+g\|_p\le\|f\|_p+\|g\|_p
$$
We can repeatedly apply this inequality to get
$$
\left\|\sum_k\;f_k\right\|_p\le\sum_k\|f_k\|_p
$$
Then using the scaling property of $\|\cdot\|_p$, we can write
$$
\left\|\sum_k\;f_k\Delta_k\right\|_p\le\sum_k\|f_k\|_p\Delta_k
$$
So we can write a Riemann sum for the integral inequality you cite and pass to the limit to get the integral inequality you cite.
Another way to show this inequality is to compute $\|\cdot\|_p$ using duality.  That is,
$$
\begin{align}
\left\|\int f(\cdot,y)\;\mathrm{d}y\right\|_p&=\sup_{\|h\|_{L^q}=1}\int\int h(x)f(x,y)\;\mathrm{d}y\;\mathrm{d}x\tag{1}\\
\int\|f(.,y)\|_p\;\mathrm{d}y&=\int\sup_{\|h\|_{L^q}=1}\int h(x)f(x,y)\;\mathrm{d}x\;\mathrm{d}y\tag{2}
\end{align}
$$
where $\frac{1}{p}+\frac{1}{q}=1$. Since any $h$ used in $(1)$ can be used for all $y$ in $(2)$, it is clear that $(2)$ is at least as big as $(1)$.  However, we might be able to do better in $(2)$ by choosing a different $h$ for each $y$.  Therefore,
$$
\begin{align}
\left\|\int f(\cdot,y)\;\mathrm{d}y\right\|_p&=\sup_{\|h\|_{L^q}=1}\int\int h(x)f(x,y)\;\mathrm{d}y\;\mathrm{d}x\\
&\le\int\sup_{\|h\|_{L^q}=1}\int h(x)f(x,y)\;\mathrm{d}x\;\mathrm{d}y\\
&=\int\|f(.,y)\|_p\;\mathrm{d}y
\end{align}
$$
