[proofplan]
We prove Minkowski's integral inequality in three cases. The case $p = 1$ is Fubini--Tonelli (the two sides are literally equal). The case $p = \infty$ is a pointwise bound. The main case $1 < p < \infty$ uses a duality argument: write $\|G\|_{L^p}^p = \int G^{p-1} G \, d\mu$, expand the inner $G$, apply Fubini--Tonelli to exchange integration order, then apply Hölder's inequality to the inner integral with exponents $p$ and $p'$. After simplification, dividing both sides by $\|G\|_{L^p}^{p-1}$ yields the result.
[/proofplan]
[step:Handle the case $p = 1$ via Fubini--Tonelli]
By Fubini--Tonelli (the integrand is non-negative, so the interchange is justified without integrability hypotheses),
\begin{align*}
\int_X \int_Y |F(x,y)| \, d\nu(y) \, d\mu(x) &= \int_Y \int_X |F(x,y)| \, d\mu(x) \, d\nu(y).
\end{align*}
The left side equals $\left\|\int_Y |F(\cdot,y)| \, d\nu(y)\right\|_{L^1(\mu)}$, and the right side equals $\int_Y \|F(\cdot,y)\|_{L^1(\mu)} \, d\nu(y)$.
The two sides are equal, so the inequality holds (with equality) for $p = 1$.
[/step]
[step:Handle the case $p = \infty$ via a pointwise bound]
For $\mu$-a.e. $x \in X$,
\begin{align*}
\left|\int_Y F(x,y) \, d\nu(y)\right| &\le \int_Y |F(x,y)| \, d\nu(y) \le \int_Y \|F(\cdot,y)\|_{L^\infty(\mu)} \, d\nu(y),
\end{align*}
which is a constant independent of $x$.
Taking the essential supremum over $x$ gives the result for $p = \infty$.
[/step]
[step:Prove the case $1 < p < \infty$ by a duality argument]
Define $G: X \to [0,\infty]$ by $G(x) := \int_Y |F(x,y)| \, d\nu(y)$.
We must show $\|G\|_{L^p(\mu)} \le \int_Y \|F(\cdot,y)\|_{L^p(\mu)} \, d\nu(y)$.
Write
\begin{align*}
\|G\|_{L^p}^p &= \int_X G(x)^p \, d\mu(x) = \int_X G(x)^{p-1}\,G(x) \, d\mu(x).
\end{align*}
Expanding $G(x)$ in the last factor:
\begin{align*}
\|G\|_{L^p}^p &= \int_X G(x)^{p-1} \int_Y |F(x,y)| \, d\nu(y) \, d\mu(x).
\end{align*}
By Fubini--Tonelli (the integrand is non-negative):
\begin{align*}
\|G\|_{L^p}^p &= \int_Y \int_X G(x)^{p-1}\,|F(x,y)| \, d\mu(x) \, d\nu(y).
\end{align*}
Apply [Hölder's inequality](/theorems/516) to the inner integral with exponents $p$ and $p' = p/(p-1)$.
The two functions being paired are $|F(\cdot,y)| \in L^p(\mu)$ and $G^{p-1} \in L^{p'}(\mu)$:
\begin{align*}
\int_X G(x)^{p-1}\,|F(x,y)| \, d\mu(x) &\le \|G^{p-1}\|_{L^{p'}(\mu)}\,\|F(\cdot,y)\|_{L^p(\mu)}.
\end{align*}
Since $(p-1)p' = (p-1) \cdot p/(p-1) = p$, we have $\|G^{p-1}\|_{L^{p'}} = \left(\int G^{(p-1)p'} \, d\mu\right)^{1/p'} = \left(\int G^p \, d\mu\right)^{1/p'} = \|G\|_{L^p}^{p/p'} = \|G\|_{L^p}^{p-1}$.
Substituting back:
\begin{align*}
\|G\|_{L^p}^p &\le \|G\|_{L^p}^{p-1} \int_Y \|F(\cdot,y)\|_{L^p(\mu)} \, d\nu(y).
\end{align*}
If $\|G\|_{L^p} = 0$ the result holds.
Otherwise, divide both sides by $\|G\|_{L^p}^{p-1}$ to obtain
\begin{align*}
\|G\|_{L^p} &\le \int_Y \|F(\cdot,y)\|_{L^p(\mu)} \, d\nu(y).
\end{align*}
[guided]
The duality trick works as follows. We want to bound $\|G\|_{L^p}$, where $G(x) = \int_Y |F(x,y)| \, d\nu(y)$. The idea is to test $G$ against itself raised to the power $p-1$: we multiply $G$ by $G^{p-1}$ and integrate, obtaining $\|G\|_{L^p}^p$. Expanding the inner copy of $G$ and applying Fubini--Tonelli (justified since the integrand $G(x)^{p-1}|F(x,y)|$ is non-negative):
\begin{align*}
\|G\|_{L^p}^p &= \int_Y \int_X G(x)^{p-1}\,|F(x,y)| \, d\mu(x) \, d\nu(y).
\end{align*}
Now apply [Hölder's inequality](/theorems/516) to the inner integral. The exponents are $p$ (applied to $|F(\cdot,y)|$) and $p' = p/(p-1)$ (applied to $G^{p-1}$). Why $p'$? Because we need $G^{p-1}$ to belong to $L^{p'}$, which requires $\int G^{(p-1)p'} \, d\mu < \infty$. Since $(p-1)p' = p$, this is exactly $\int G^p \, d\mu = \|G\|_{L^p}^p$, which we are trying to bound. Hölder gives:
\begin{align*}
\int_X G(x)^{p-1}\,|F(x,y)| \, d\mu(x) &\le \|G\|_{L^p}^{p-1}\,\|F(\cdot,y)\|_{L^p(\mu)}.
\end{align*}
Integrating over $y$:
\begin{align*}
\|G\|_{L^p}^p &\le \|G\|_{L^p}^{p-1} \int_Y \|F(\cdot,y)\|_{L^p(\mu)} \, d\nu(y).
\end{align*}
If $\|G\|_{L^p} = 0$, the inequality $\|G\|_{L^p} \le \int \|F(\cdot,y)\|_{L^p} \, d\nu(y)$ is immediate. If $\|G\|_{L^p} > 0$, dividing both sides by $\|G\|_{L^p}^{p-1}$ (which is finite and positive) yields:
\begin{align*}
\|G\|_{L^p} &\le \int_Y \|F(\cdot,y)\|_{L^p(\mu)} \, d\nu(y).
\end{align*}
The division requires $\|G\|_{L^p}^{p-1} < \infty$, which holds because $\|G\|_{L^p}^p \le \|G\|_{L^p}^{p-1} \cdot (\text{finite RHS})$ forces $\|G\|_{L^p} < \infty$ (if $\|G\|_{L^p} = \infty$ the right side of the Hölder bound would also be $\infty$, but the RHS of the original inequality is assumed finite).
[/guided]
[/step]
