I am reading the proof of the Sobolev embedding theorem in Brezis' book, but i can't understand the following fact.
Suppose we've already proved the following lemma:
Let $f_1$, $f_2$, $\dots$, $f_N \in L^{N-1}(\mathbb{R}^{N-1})$ and set $$\tilde{x_i}=(x_1,x_2,\dots,x_{i-1},x_{i+1},\dots,x_N) \in \mathbb{R}^{N-1}.$$ Then the function $$ f(x)=f_1(\tilde{x_1})\dots f_N(\tilde{x_N}), \quad x \in\mathbb{R}^N$$ is in $L^{1}(\mathbb{R}^N)$ and $$||f||_{L^{1}(\mathbb{R}^N)} \leq \prod_{i=1}^{N}||f_i||_{L^{N-1}(\mathbb{R}^{N-1})}.$$
Now, in the theorem, one considers $u\in C_c^1(\mathbb{R}^N)$ and $$ f_i(\tilde{x_i}):= \int_{\mathbb{R}}\bigg|\frac{\partial u}{\partial x_i}(x_1,x_2,\dots,x_{i-1},t,x_{i+1},\dots,x_N)\bigg|dt $$ so that $$ |u(x)|^N \leq \prod_{i} f_i(\tilde{x_i}) $$ and then one should be able (by using the previous lemma) - and this is the point I am missing - to prove that $$ \int_{\mathbb{R}^N} |u(x)|^{\frac{N}{N-1}}dx \leq \prod_{i} ||f_i||^\frac{1}{N-1}_{L^1(\mathbb{R}^{N-1})} $$
Raise both sides of your penultimate inequality to the power $1/(N-1)$ and integrate to get $$ \int_{\mathbb{R}^N} |u(x)|^{N/(N-1)}dx \le \int_{\mathbb{R}^N} \prod_{i} f_i^{1/(N-1)}. $$ Now apply your lemma to $$ g = \prod_{i} f_i^{1/(N-1)} $$ with $g_i = f_i^{1/(N-1)}$. This gives $$ \int_{\mathbb{R}^N} \prod_{i} f_i^{1/(N-1)} = \Vert g \Vert_{L^1(\mathbb{R}^N)} \le \prod_{i=1}^N \Vert g_i \Vert_{L^{N-1}(\mathbb{R}^{N-1})} = \prod_{i=1}^N \Vert f_i \Vert_{L^{1}(\mathbb{R}^{N-1})}^{1/(N-1)}. $$ Hence $$ \int_{\mathbb{R}^N} |u(x)|^{N/(N-1)} dx\le \prod_{i=1}^N \Vert f_i \Vert_{L^{1}(\mathbb{R}^{N-1})}^{1/(N-1)}, $$ which is what you were after.