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For $x_i\ge0$, where $i=1,2,...,n$, satisfying

$\sum_{i=1}^n\,x_i^2+2\,\sum_{1\le k

find the maximum and minimum of $\sum_{i=1}^n x_i.$

2 Answers 2

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Actually the minimum is pretty trivial:

$\left(\sum_{i=1}^n\,x_i\right)^2\ge \sum_{i=1}^{n}\,x_{i}^{2}+2\,\sum_{1\le k For equality can we take $x_1=1$ and $x_i=0$ for $i>1$.

The maximum is much more interesting.

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Let $y_i:=\frac{x_i}{\sqrt{i}}$ for every $i=1,2,\ldots,n$. Then, $1=\sum_{i=1}^n\,x_i^2+2\,\sum_{1\leq k Define $z_i:=\sum_{j=i}^n\,y_j$ for all $i=1,2,\ldots,n$. We have $\sum_{i=1}^n\,z_i^2=1$ and $\sum_{i=1}^n\,x_i=\sum_{i=1}^n\,\sqrt{i}\,y_i=\sum_{i=1}^n\,\left(\sqrt{i}-\sqrt{i-1}\right)\,z_i\,.$ Thus, by the Cauchy-Schwarz Inequality, we have $\sum_{i=1}^n\,\left(\sqrt{i}-\sqrt{i-1}\right)\,z_i\leq\sqrt{\sum_{i=1}^n\,z_i^2}\sqrt{\sum_{i=1}^n\,\left(\sqrt{i}-\sqrt{i-1}\right)^2}=\sqrt{n^2-2\,\sum_{i=1}^n\,\sqrt{i(i-1)}}\,.$ The equality holds if and only if $z_i=\frac{\sqrt{i}-\sqrt{i-1}}{\lambda}$ for every $i=1,2,\ldots,n$, where $\lambda:=\sqrt{n^2-2\,\sum_{i=1}^n\,\sqrt{i(i-1)}}\,.$ Hence, the maximum value of $\sum_{i=1}^n\,x_i$ is $\lambda$, which occurs if and only if

(1) $y_i=\frac{2\sqrt{i}-\sqrt{i+1}-\sqrt{i-1}}{\lambda}$, or $x_i=\frac{2i-\sqrt{i(i+1)}-\sqrt{i(i-1)}}{\lambda}$ for all $i=1,2,\ldots,n-1$, and

(2) $y_n=\frac{\sqrt{n}-\sqrt{n-1}}{\lambda}$, or equivalently, $x_n=\frac{n-\sqrt{n(n-1)}}{\lambda}$.