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An inequality from the following paper

MIKOLAS, M., Sur un probleme d'extremum et une extension de l'inegalite de Minkowski, Ann. Univ. Sci. Budapest Eotvos, Sect. Math. 1 (1958), 101-106

is discussed in Mitrinovic's Analytic Inequalities book. The inequality is stated as below.

Let $x_{uv},$ $p,r$ be positive numbers with $\sum_{u=1}^m x_{uv}=c_v.$ Then $$ \sum_{u=1}^m \left(\sum_{v=1}^n x_{uv}^r\right)^p \leq n^{1-pr} \left(\sum_{v=1}^n c_v^r \right)^p $$ if $r\leq 1,$ and $pr\leq 1.$ The converse inequality holds if $r\geq 1,$ and $pr\geq 1.$

I have no access to the paper, so I was wondering

Question: Is the positivity of the matrix entries necessary for this inequality to hold?

To avoid trivialities, we may assume each row/column has at least one nonzero member.

However, it does seem like if we let $r\rightarrow 0,$ then problems may arise since the inner sums on the LHS would tend to the support size of the row sums and some high support size rows can spoil the inequality.

So, maybe if we assume some lower bound on $r$, like $r>1/2,$ it might be possible to recover the inequality, even with some zero entries.

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As suggested by @FrancoisZiegler, I looked at the original with my basic French and have concluded a few things.

Erratum: The theorem is wrongly stated in Mitrinovic, as well as the more recent Mitrinovic, Pecaric and Fink Classical and New Inequalities in Analysis, Chapter 8, p. 107. The correct statement from the original paper is below, namely the term $n^{1-pr}$ should be $m^{1-pr}$ on the right hand side.

Let $x_{uv},$ $p,r$ be positive numbers with $\sum_{u=1}^m x_{uv}=c_v.$ Then $$ \sum_{u=1}^m \left(\sum_{v=1}^n x_{uv}^r\right)^p \leq m^{1-pr} \left(\sum_{v=1}^n c_v^r \right)^p $$ if $r\leq 1,$ and $pr\leq 1.$ The converse inequality holds if $r\geq 1,$ and $pr\geq 1.$

It is also proved in Mikolas' original paper that the positivity of the entries $x_{uv}$ is needed for the result to hold.

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