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I'm not sure how to solve the following problem: $$d_1^2 + \ldots + d_n^2 = \sigma^2$$ $$d_i \geq 0$$

Find the minimum possible value of $$d_1 + \ldots + d_n$$

I have a hunch that its when every value is zero except one of the $d_i$ values. This holds for n = 1, n = 2, and n = 3 (confirmed with 3d graphing calc). However, I'm not sure how to prove it for a general n.

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    $\begingroup$ Do you know the Langrange multiplier method? $\endgroup$ – amsmath Apr 19 at 3:56
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    $\begingroup$ By Lagrange's method of undetermined multiplier we have $\text {Max}\ (d_1+ d_2 + \cdots + d_n)$ subject to $d_1^2+ d_2^2 + \cdots + d_n^2 = {\sigma}^2$ is $ {\sigma} {\sqrt n}.$ $\endgroup$ – Dbchatto67 Apr 19 at 4:02
  • $\begingroup$ Just to clarify, Lagrange multipliers is not failing here, it's just not useful. If the problem did not have the constraint of all terms being non-negative, the method would correctly give the respective maximum and minimum as $\sigma \sqrt n, - \sigma \sqrt n$. But with that additional constraint the only extremum the method is able to identify is the maximum. The minimum needs another method, as already given here. $\endgroup$ – Deepak Apr 19 at 5:05
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Observe that $$d_1^2+d_2^2 + \cdots + d_n^2 =\left ( \sum\limits_{i=1}^{n} d_i \right )^2 - 2 \sum\limits_{i<j} d_id_j.$$ So we have $$\begin{align*} \left ( \sum\limits_{i=1}^{n} d_i \right )^2 & = d_1^2+d_2^2 + \cdots + d_n^2 + 2\sum\limits_{i<j} d_id_j. \\ & \geq d_1^2+d_2^2 + \cdots + d_n^2 = {\sigma}^2. \end{align*}$$

The last inequality holds since $d_i \geq 0$ for all $i=1,2, \cdots , n.$

So $$\sum\limits_{i=1}^{n} d_i \geq \sigma.$$ But if we choose $d_i=\sigma$ and $d_j =0$ for all $j \neq i$ then the constraint $d_1^2 + d_2 ^2 + \cdots + d_n^2 = {\sigma}^2$ is satisfied. But then we have $\sum\limits_{i=1}^{n} d_i = \sigma.$ So we can conclude that $$\text {Min} \left (\sum\limits_{i=1}^{n} d_i \right ) = \sigma$$ subject to the constraint $\sum\limits_{i=1}^{n} d_i^2 = {\sigma}^2.$

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