Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$$\sum_{i=1}^n\,x_i^2+2\,\sum_{1\le k <j\le n}\,\sqrt{\frac{k}{j}}\,x_kx_j=1\,,$$

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

share|cite|improve this question

Let $y_i:=\frac{x_i}{\sqrt{i}}$ for every $i=1,2,\ldots,n$. Then, $$1=\sum_{i=1}^n\,x_i^2+2\,\sum_{1\leq k<j\leq n}\,\sqrt{\frac{k}{j}}\,x_kx_j=\sum_{i=1}^n\,i\,y_i^2+2\sum_{1\leq k < j \leq n}\,k\,y_ky_j=\sum_{i=1}^n\left(\sum_{j=i}^n\,y_j\right)^2\,.$$ 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, $x_n=\frac{n-\sqrt{n(n-1)}}{\lambda}$.

share|cite|improve this answer

Actually the minimum is pretty trivial:

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

The maximum is much more interesting.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.