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Let $a_1,a_2,\ldots,a_n$ are $n$ non-negative real numbers and for some $p>1,q>1$ with $1/p+1/q=1,$ such that $$ n\left(\sum_{j=1}^{n}(a_j)^p\right)^{q/p}<1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$ Does this imply $$ n\sum_{j=1}^{n}a_j\leq 1? $$ I tried as following: \begin{align} &n\left(\sum_{j=1}^{n}(a_j)^p\right)^{q/p}<1 ~\Rightarrow~ \sum_{j=1}^{n}(a_j)^p<\frac{1}{n^{p/q}}~{\color{Red}\Rightarrow}~ (a_j)^p<\frac{1}{n^{p/q}}\\ &\Rightarrow a_j<\frac{1}{n^{1/q}}~\Rightarrow~ \sum_{j=1}^{n}a_j=n^{1/p} \end{align} I think I loose too much at red implication. Kindly help in proving or providing counter example.

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  • $\begingroup$ but your sum is in breckets and the $n$ stands bevor the sum $\endgroup$ – Dr. Sonnhard Graubner Mar 3 '17 at 14:08
  • $\begingroup$ @MichaelRozenberg $a_1,a_2,\ldots, a_n$ satisty inequality (1). It don't think your choice of $a_j's$ satisfy (1). $\endgroup$ – Suhail Mar 3 '17 at 15:03
  • $\begingroup$ @Suhail Now I understood you. See my solution. $\endgroup$ – Michael Rozenberg Mar 3 '17 at 15:47
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Let $\frac{1}{p}=\alpha$ and $\frac{1}{q}=\beta$.

Hence, $\alpha+\beta=1$ and by P-M we obtain: $$1>n^{\frac{1}{q}}\left(\sum_{j=1}^{n}a_j^p\right)^{\frac{1}{p}}=n^{\beta}\left(\sum_{j=1}^{n}a_j^\frac{1}{\alpha}\right)^{\alpha}=n^{\alpha+\beta}\left(\frac{\sum\limits_{j=1}^{n}a_j^\frac{1}{\alpha}}{n}\right)^{\alpha}=$$ $$=n\left(\frac{\sum\limits_{j=1}^{n}a_j^\frac{1}{\alpha}}{n}\right)^{\alpha}\geq n\cdot\frac{\sum\limits_{j=1}^na_j}{n}=\sum\limits_{j=1}^na_j$$

A counterexample is $a_1=a_2=...=a_n\rightarrow\frac{1}{n}$.

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