# Equality case in elementary form of Holder's Inequality

A well known elementary formulation of Holder's Inequality can be stated as follows:

Let $a_{ij}$ for $i = 1, 2, \dots, k; j = 1, 2, \dots, n$ be positive real numbers, and let $p_1, p_2, \dots, p_k$ be positive real numbers such that $\sum_{i=1}^{k} \frac{1}{p_i} = 1$. Then we have

$$\sum_{j = 1}^{n} \prod_{i = 1}^{k} a_{ij} \leq \prod_{i=1}^{k} \left ( \sum_{j = 1}^{n} a_{ij}^{p_i} \right )^{\frac{1}{p_i}}.$$

Does anyone know when equality occurs in this inequality? Any insights on the equality case and/or a proof would be appreciated.

I have seen several proofs of the two sequence case. Here is one, from Cvetkovski's Inequalities:

Perhaps this argument can be generalized to prove the above result. Thanks in advance!

Since the usual proof of generalized Holder's proceeds by induction from the usual Holder's inequality, the equality case is precisely the same. In other words, if $v_j$ denotes the vector with components $a_{ij}^p$, then equality occurs iff all the vectors $v_j$ are parallel.