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If $p \geqslant 1$, and let $c^p \leqslant \sum\limits_{i=1}^n a_i^p $ . Then $$ c \leqslant \sum_{i=1}^n a_i$$ Here $c , a_i \geqslant 0$.

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$\|a\|_p = \left( \sum\limits_{i=1}^n a_i^p\right)^{1/p}$ is the $\ell_p$ norm of $(a_1,\ldots,a_n)$. The fact that this is subadditive, and thus a norm, is Minkowski's inequality. Thus $$\|a\|_p \le \|(a_1,0,\ldots,0)\|_p + \|(0,a_2,0,\ldots,0)\|_p + \ldots + \|(0,\ldots,0,a_n)\|_p = \sum_{i=1}^n |a_i|$$

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An elementary argument might go as follows: The case $c=0$ is clear. Thus, assume $c\neq 0.$ We can reduce to the case $c=1$ by dividing through by $c^p$ in the first inequality, and by $c$ in the second inequality. So suppose that $\sum a_i^p \ge 1.$ If all $a_i< 1,$ then $1\le \sum a_i^p \le \sum a_i$ is clear. Otherwise there is some $a_i \ge 1$ and again we are done.

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