A discrete variant, where we consider a finite subset of the 3-dimensional space, was asked at the IMO in 1992. See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60719. The IMO problem gives us some hint how prove it in any dimension, and the proofs can be extended to functions or measures.
For the inductive proof of the discrete problem, the key is to find the case of equality. Obviously, equality holds if the set is of the form $A\times B\times C$ with finite sets $A,B,C\subset\mathbb{R}$. For the induction step, you can split one of the sets in two smaller subsets, apply the induction hypotheses and then apply Cauchy or Hölder to complete.
(I have seen this continuous form too, but could not find it.)
An inductive solution, using Hölder is this. For $n=2$ the statement is trivial:
$$ \int_x \int_y f(y) f(x) = \left(\int_y f(y)\right) \cdot \left(\int_x g(x)\right) = I_1I_2. $$
Now suppose that $n\ge 3$ and the statement holds true for $n-1$.
Apply Hölder twice as
\begin{gather*}
\int_{x_1}\ldots\int_{x_n} f_1\ldots f_n =
\int_{x_1}\ldots\int_{x_{n-1}} \left( f_n \int_{x_n} f_1\dots f_{n-1} \right)
\le \\ \le
\left(\int_{x_1}\ldots\int_{x_{n-1}} f_n^{n-1} \right)^{\frac1{n-1}}
\left(\int_{x_1}\ldots\int_{x_{n-1}} \left(\int_{x_n} f_1\dots f_{n-1}\right)^{\frac{n-1}{n-2}}\right)^{\frac{n-2}{n-1}}
\le \\ \le
I_n^{\frac1{n-1}}
\left(\int_{x_1}\ldots\int_{x_{n-1}} \prod_{i=1}^{n-1} \left(\int_{x_n} f_i^{n-1}\right)^{\frac1{n-2}}\right)^{\frac{n-2}{n-1}}.
\end{gather*}
Applying the induction hypothesis to the functions
$F_i(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{n-1})=\left(\int_{x_n}f_i^{n-1}(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{n-1},x_n)\mathrm{d}x_n\right)^{\frac1{n-2}}$, we get
\begin{gather*}
\int_{x_1}\ldots\int_{x_{n-1}} \prod_{i=1}^{n-1} \left(\int_{x_n} f_i^{n-1}\right)^{\frac1{n-2}} =
\int_{x_1}\ldots\int_{x_{n-1}} F_1\ldots F_{n-1}
\le\\\le
\prod_{i=1}^{n-1} \left( \int_{x_1}\ldots\int_{x_{i-1}}\int_{x_{i+1}}\ldots\int_{x_{n-1}} F_i^{n-2} \right)^{\frac1{n-2}}
= \prod_{i=1}^{n-1} \left( \int_{x_1}\ldots\int_{x_{i-1}}\int_{x_{i+1}}\ldots\int_{x_n} f_i^{n-1} \right)^{\frac1{n-2}}
=\\= (I_1\ldots I_{n-1})^{\frac1{n-2}}.
\end{gather*}
Hence,
\begin{gather*}
\int_{x_1}\ldots\int_{x_n} f_1\ldots f_n \le
I_n^{\frac1{n-1}}
\left(\int_{x_1}\ldots\int_{x_{n-1}} \prod_{i=1}^{n-1} \left(\int_{x_n} f_i^{n-1}\right)^{\frac1{n-2}}\right)^{\frac{n-2}{n-1}} \le
(I_1\ldots I_n)^{\frac1{n-1}}.
\end{gather*}
Done.
