Problem: Let $f$ be a measurable nonnegative function on $[0,1]^2$, and $1\leq r < p < \infty$. Then, show that $$ \left(\int_{0}^{1}\left(\int_{0}^{1}f^{r}(x,y)dy\right)^{p/r} \right)^{1/p} \leq \left(\int_{0}^{1}\left(\int_{0}^{1}f^{p}(x,y)dy\right)^{r/p} \right)^{1/r}.$$ Hint : Let $s=\frac{p}{r}$ and $F(x) = \int_{0}^{1}f^{r}(x,y)dy$ and consider appropriate function $h \in L_{s'}[0,1].$
My attempt: By following hints, I've got this result. For any $h \in L_{s'}[0,1],$ from the Tonelli's theorem,
$$ \int_{0}^{1}F(x)|h(x)|dx = \int_{0}^{1}\int_{0}^{1}f^{r}(x,y)dy|h(x)|dx = \int_{0}^{1}\int_{0}^{1}f^{r}(x,y)|h(x)|dxdy, $$ and by using the Holder's inequality, $$ \int_{0}^{1}\int_{0}^{1}f^{r}(x,y)|h(x)|dxdy \leq ||h||_{s'}\int_{0}^{1}||f^{r}(x,y)||_{s} = ||h||_{s'}\int_{0}^{1}\left(\int_{0}^{1}f^{p}(x,y)dx \right)^{r/p}dy.$$
However, now I have no idea how to proceed. I know that $$||F||_{s}^{1/r} = \left(\int_{0}^{1}\left(\int_{0}^{1}f^{r}(x,y)dy\right)^{p/r} \right)^{1/p} $$ so I tried to choose suitable $h \in L_{s'}[0,1]$ with $||h||_{L_{s'}[0,1]}$ such that $$\int_{0}^{1}F(x)|h(x)|dx =||F||_{s}.$$ If we find such function, than $$||F||_{s} \leq \int_{0}^{1}\left(\int_{0}^{1}f^{p}(x,y)dx \right)^{r/p}dy$$ so we are done by taking $1/r$-th root.
However I haven't find yet. If you gave me one more step for this problem, I would appreciate much about it.
Hint: choose $h(x)=F(x)^a\left\lVert F\right\rVert_s^b$ for some well chosen parameters $a$ and $b$.