Prove generalised Hölder's inequality without calculus or analysis.

Question

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 $m=n=3$

Answer 1

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.