Generalised Holder ineq Prove the following generalisation of Holder's inequality
$$\int | u_1 \cdot ... \cdot u_N | d\mu \leq \|u_1\|_{p_1} \cdot ... \cdot \|u_N\|_{p_N}$$
for all $p_j \in (1,\infty)$ such that $\sum_{j=1}^N p_j^{-1} = 1$ and all measurable $u_j$.
Proof by induction
$\bullet$ The base case is easy
$\bullet$ Now we suppose the inequality holds for some $N\in \mathbb{N}$. Let $p_1,...,p_N, q >1$ be numbers such that $\underbrace{\sum_{j=1}^N p_j^{-1}}_{P} + q^{-1} = 1$.
Applying the regular Holder inequality we get
$$\int | u_1 \cdot ... \cdot u_N \cdot w| \,\,d\mu \leq \left( \int | u_1 \cdot ... \cdot u_N |^P \, d\mu \right)^{1/P} \cdot \left( \int |w| \, d\mu \right)^{1/q}$$
Now we can use our induction hypothesis for the first factor on the RHS, but how?? I've tried using it for $\alpha_j = \frac{P}{p_j}$, $\sum_{j=1}^N a_j^{-1} = 1$ but it doesn't seem to yield the desired result.
I'm stuck, please help.
 A: Inductively on $n$.
Assume that, the Gen. Holder holds for $n=k$. That is, for every $p_1,\ldots,p_k\in(1,\infty)$, with $\sum_{j=1}^k p_j^{-1}=1$, we have 
$$
\int |u_1\cdots u_k|\le \|u_1\|_{p_1}\cdots\|u_k\|_{p_k},
$$
and let $q_1,\ldots,q_k,q_{k+1}\in (1,\infty)$, with $\sum_{j=1}^{k+1} q_j^{-1}=1$. Then
$$
\sum_{j=1}^{k} q_j^{-1}=1-q^{-1}_{k+1}=\frac{q_{k+1}-1}{q_{k+1}},
$$
and 
$$
\int\lvert
u_1\cdots u_ku_{k+1}\rvert\le\|u_1\cdots u_k\|_{q_{k+1}^*}\cdot\|u_{k+1}\|_{q_{k+1}},
$$
where $q_{k+1}^*=q_{k+1}/(q_{k+1}-1)$. But
$$
\|u_1\cdots u_k\|_{q_{k+1}^*}=\Big(\int \lvert u_1\cdots u_k\rvert^{q_{k+1}^*}\Big)^{1/q_{k+1}^*}
=\Big(\int 
\lvert u_1\rvert^{\frac{q_{k+1}}{q_{k+1}-1}}\cdots \lvert u_k\rvert^{\frac{q_{k+1}}{q_{k+1}-1}}
\Big)^{\frac{q_{k+1}-1}{q_{k+1}}}\tag{1}
$$
Set now $w_j=\lvert u_j\rvert^{\frac{q_{k+1}}{q_{k+1}-1}}$. Then
$$
\Big(\int 
\lvert u_1\rvert^{\frac{q_{k+1}}{q_{k+1}-1}}\cdots \lvert u_k\rvert^{\frac{q_{k+1}}{q_{k+1}-1}}
\Big)^{\frac{q_{k+1}-1}{q_{k+1}}}=\Big(\int 
\lvert w_1\cdots w_k\rvert
\Big)^{\frac{q_{k+1}-1}{q_{k+1}}}\le
\Big(
\| w_1\|_{\hat q_i}\cdots \|w_k\|_{\hat q_k}
\Big)^{\frac{q_{k+1}-1}{q_{k+1}}},\tag{2}
$$
where 
$$
\hat q_j=q_j\cdot\frac{q_{k+1}-1}{q_{k+1}},
$$
as
$$
\sum_{j=1}^k \hat q_j^{-1}=\sum_{j=1}^k q_j^{-1}\cdot\frac{q_{k+1}}{q_{k+1}-1}=1.
$$
Now, 
$$
\|w_j\|_{\hat q_j}=\Big(\int\lvert w_j\rvert^{\hat q_j}\Big)^{1/\hat q_j}
=\Big(\int\lvert u_j\rvert^{q_{k+1}\hat q_j/(q_{k+1}-1)}\Big)^{1/\hat q_j}
=\Big(\int\lvert u_j\rvert^{q_j}\Big)^{1/\hat q_j}=\|u_j\|_{q_j}^{q_{k+1}/(q_{k+1}-1)}
\tag{3}$$
Combining $(1)-(3)$ we complete the inductive hypothesis.
