= Solution
For <conjugate exponents> $p,q\in[1,\infty]$, $1/p+1/q=1$, <Hölder's inequality> states that $\int|fg|\leq\|f\|_p\|g\|_q$. If $r=\infty$ in the requested product estimate, every $p_i=\infty$ and multiplication of essential bounds proves it directly. Suppose $r<\infty$. Factors with $p_i=\infty$ can again be bounded pointwise by their <essential suprema>.
Here is an explicit finite-exponent proof of the remaining <Generalized Holder inequality>. Put $q_i=p_i/r$, so $\sum_i1/q_i=1$, and normalize $h_i=|f_i|^r/\|f_i\|_{p_i}^r$. If one <norm> is zero the product vanishes <almost everywhere>. Otherwise $\int h_i^{q_i}=1$. The weighted <arithmetic-geometric mean inequality>, applied to $h_i^{q_i}$ with weights $1/q_i$, gives
$$
\prod_i h_i\leq\sum_i\frac{h_i^{q_i}}{q_i}.
$$
Integrating gives $\int\prod_i h_i\leq1$. If only one finite exponent remains, its weight is one and the same argument is an equality. Restoring the normalization and taking the $r$th root proves
$$
\boxed{\|f_1\cdots f_n\|_r\leq\prod_{i=1}^n\|f_i\|_{p_i}.}
$$
In particular the product belongs to $L^r$.
For the projection estimate, use the three-dimensional <functional Loomis-Whitney inequality>
$$
\int_{\mathbb R^3}
\bigl(g_1(x_2,x_3)g_2(x_1,x_3)g_3(x_1,x_2)\bigr)^{1/2}\,dx
\leq\prod_{j=1}^3\left(\int_{\mathbb R^2}g_j\right)^{1/2}.
$$
This permitted three-dimensional result also follows directly by <Cauchy-Schwarz inequality>: integrate the <square root> of $g_1g_2$ in $x_3$, obtaining the <square root> of the product of their marginal <integrals>, and then apply <Cauchy-Schwarz inequality> in $(x_1,x_2)$ against $\sqrt{g_3}$.
Let $f_j$ be any nonnegative <integrable function> of the three coordinates excluding $x_j$. For $j=1,2,3$, define $G_j$ by integrating $f_j$ in $x_4$. First <Hölder's inequality> with three equal exponents yields
$$
\int_{\mathbb R}\prod_{j=1}^3 f_j(\widehat x_j)^{1/3}\,dx_4
\leq\prod_{j=1}^3G_j^{1/3}.
$$
The fourth factor $f_4(x_1,x_2,x_3)$ is independent of $x_4$. A further <Hölder's inequality> with exponents $3$ and $3/2$, followed by the three-dimensional estimate, therefore gives
$$
\begin{aligned}
\int_{\mathbb R^4}\prod_{j=1}^4f_j(\widehat x_j)^{1/3}\,dx
&\leq\left(\int_{\mathbb R^3}f_4\right)^{1/3}
\left(\int_{\mathbb R^3}\prod_{j=1}^3G_j^{1/2}\right)^{2/3}\\
&\leq\prod_{j=1}^4\left(\int_{\mathbb R^3}f_j\right)^{1/3}.
\end{aligned}
$$
The marginal-integral identifications use <Tonelli's theorem>. Now take $f_j=\mathbf1_{K_j}$. The projections are compact and hence measurable with <finite measure>. Every $x\in K$ has all its projected coordinates in the corresponding $K_j$, so $\mathbf1_K(x)\leq\prod_j\mathbf1_{K_j}(\widehat x_j)^{1/3}$. Consequently
$$
\boxed{\lambda_4(K)\leq\left(\prod_{j=1}^4\lambda_3(K_j)\right)^{1/3}.}
$$
This includes zero-measure projections and is sharp for axis-parallel rectangular boxes.
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