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I would like to show that $$\Big(\int_X \Big(\int_Y |f(x,y)|^qdy\Big)^\frac{p}{q}dx\Big)^\frac{1}{p} \leq \Big(\int_Y \Big(\int_X |f(x,y)|^pdx\Big)^\frac{q}{p}dy\Big)^\frac{1}{q}$$ for $0 < q \leq p < \infty.$

For each fixed $x, y$, let $f_y(x) = f(x,y)$ and $f_x(y) = f(x,y).$ Then the inequality says $$\Big|\Big|||f_x||_q\Big|\Big|_p \leq \Big|\Big|||f_y||_p\Big|\Big|_q.$$ So it reminds me of mixing Fubini Theorem with Holder inequality might help. But $p, q$ here does not even in $(1, \infty)$ and does not a conjugate exponent. Also, the function change form $f_x$ to $f_y$. So I do not quite sure where to begin.

Both Htob
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Let $r>0$ be such that $\dfrac{1}{r}+\dfrac{1}{p/q}=1$, denoting \begin{align*} I(x)=\int_{Y}|f(x,y)|^{q}dy, \end{align*} then \begin{align*} \int_{X}\left(\int_{Y}|f(x,y)|^{q}dy\right)^{p/q}dx&=\int_{X}I(x)\cdot I(x)^{(p/q)-1}dx\\ &= \int_{X}\left(\int_{Y} |f(x,y)|^{q} \cdot|f(x,y)|^{(p/q -1)}dy\right)dx \text{, by Fubini's Theorem} \\ &= \int_{Y} \int_{X} |f(x,y)|^q \cdot|f(x,y)|^{(p/q -1)}dx dy \\ &=\int_{Y}\int_{X}|f(x,y)|^{q}I(x)^{(q/p)-1}dxdy \text{ , by Holder's inequality}\\ &\leq\int_{Y}\left(\int_{X}(|f(x,y)|^{q})^{p/q}dx\right)^{q/p}\left(\int_{X}(I(x)^{(q/p)-1})^{r}dx\right)^{1/r}dy, \end{align*} after simplifying, the result follows.

user770687
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user284331
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