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If $\sum_{i = 1}^n a_i = x$ for $a_i \geq 0$, then is it possible to find upper bound to $\sum_{i=1}^n a_i^2$? I know that the lower bound can be easily determined using Cauchy- Schwarz Inequality. But how to derive the upper bound?

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  • $\begingroup$ Welcome to MSE. If you want to type in italic, enclose your text between a pair of asterisks. $\endgroup$ Apr 18, 2021 at 9:43

3 Answers 3

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One simple bound is $$\sum_{i=1}^{n} a_i^2 \leq \sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j = \left(\sum_{i=1}^{n} a_i\right) \left(\sum_{j=1}^{n} a_j\right) = x^2.$$ The inequality holds because the LHS is the sum of just the $i=j$ terms in the double sum.

This bound is achieved if and only if $a_i a_j = 0$ whenever $i \neq j$, which is true if and only if at most one of the $a_i$'s is nonzero.

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Do you assume that $a_i\ge 0$ ?

Otherwise take $a_1+a_2=0$ with $a_1=t$, $a_2=-t$, then $a_1^2+a_2^2=2t^2$ has no upper bound.

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  • $\begingroup$ Yes $a_i \geq 0$ $\endgroup$ Apr 18, 2021 at 9:36
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If you also know an upper bound on the $a_i \ge 0$. Then you also have: $$ \sum_{i=1}^n a_i^2 \le \max_i a_i \cdot \sum_{i=1}^n a_i = x \cdot \max_i a_i $$

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