Prove generalised Hölder's inequality without calculus or analysis. 
It is the generalised Hölder's inequality.I saw many analytical proofs in this site but I don't know analysis. So I need a basic proof. I proved it by A.M.-G.M. for $m=n=3$. Please help me.
Proof when $m=n=3$

 A: We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_{i,1}, a_{i,2}, \ldots, a_{i,n})$ for $i = 1, \ldots, m$, then
\begin{align*}
m &= \sum_{i=1}^m 1 \\
&= \sum_{i=1}^m \sum_{j=1}^n \frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} \\
1 &= \sum_{j=1}^n \frac{1}{m} \sum_{i=1}^m \frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} \\
&\ge \sum_{j=1}^n \prod_{i=1}^m \frac{\sqrt[m]{a_{i,j}}}{\sqrt[m]{\sum_{k=1}^n a_{i,k}}} \\
&= \frac{1}{\prod_{i=1}^m \sqrt[m]{\sum_{k=1}^n a_{i,k}}}\sum_{j=1}^n \sqrt[m]{\prod_{i=1}^m a_{i,j}},
\end{align*}
which implies
$$\prod_{i=1}^m \sum_{k=1}^n a_{i,k} \ge \left(\sum_{j=1}^n \sqrt[m]{\prod_{i=1}^m a_{i,j}}\right)^m$$
as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.
The inequality will be equal whenever we have
$$\frac{1}{m} \sum_{i=1}^m \frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}}
= \prod_{i=1}^m \frac{\sqrt[m]{a_{i,j}}}{\sqrt[m]{\sum_{k=1}^n a_{i,k}}}$$
for any $j = 1, \ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that
$$\frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} = C_j$$
for all $i, j$ over their respective ranges. Let $b_i = \sum_{k=1}^n a_{i,k}$. Then
$$a_{i, j} = C_j b_i,$$
in other words, the sequences $(a_{i, j})_{i=1}^m$, as $j = 1, \ldots, n$, are just multiples of each other, also as required.
