$x_{n+1}$ must be greater than $0$ because all other terms are positive. Solution: First of all we have that: $\sqrt[k] {x_i} \ge \sqrt[l] {x_i} \iff k > l$ equality holds $\iff x_i =1$ $\implies x_i \ge \sqrt{x_i}$ It implies also, that: $\sqrt{x_{n+1}} = \frac{\sum_{i=1}^{n} \sqrt{x_{i} ^{2i-1}}}{n} \ge \frac{\sum_{i=1}^{n} x_i }{n} = x_{n+1}$ $\implies x_{n+1} = \sqrt{x_{n+1}} = 1$ Now we have that: $1=\frac{\sum_{i=1}^{n} x_i}{n}\ge \frac{\sum_{i=1}^{n} 1}{n} = 1$ $\implies x_i = 1$ for $1\le i \le n+1$ And it is the only solution.

Quote: $\sqrt[k]{x_{i}}\ge \sqrt[l]{x_{i}}\iff k > l$ why?? if $x_{n+1}>1$? Quote: It implies also, that: $\sqrt{x_{n+1}}= \frac{\sum_{i=1}^{n}\sqrt{x_{i}^{2i-1}}}{n}\ge \frac{\sum_{i=1}^{n}x_{i}}{n}= x_{n+1}$ how you get $\frac{\sum_{i=1}^{n}\sqrt{x_{i}^{2i-1}}}{n}\ge \frac{\sum_{i=1}^{n}x_{i}}{n}$?

This post is really old, i didn't remeber it. For the first question it's just a typo.. The second inequality should be in the other way. Now for the second question i can't answer, there's a mistake in the way i'm reasoning. Because the problem is that $x^{a}\ge x^{b}$ if $a > b$. So $x_{i}^{\frac{2i-1}{2}}\ge x_{i}\forall i\ge 2$ but i did not consider the case $x_{1}^{\frac{1}{2}}\le x_{1}$ so there's a mistake. Sorry. I wrote this post when i was learning to use Tex so there are lots of mistakes.

OK so this does not look at all like the problem that actually came in the contest. The real problem is: Let $ n > 1$. Find all solutions for $ x_{1},x_{2},...,x_{n + 1}\geq 1$ such that: $ (1)$ $ x_{1}^{\frac {1}{2}} + x_{2}^{\frac {1}{3}} + ... + x_{n}^{\frac {1}{n + 1}} = nx_{n + 1}^{\frac {1}{2}}$; and $ (2)$ $ \frac {x_{1} + x_{2} + ... + x_{n}}{n} = x_{n + 1}$ Anyways, the answer to this problem is the following We want to find all numbers that meet $ x_{1}^{\frac {1}{2}} + ... + x_{n}^{\frac {1}{n + 1}} = n\sqrt {\frac {x_{1} + ... + x_{n}}{n}}$ By the Quadratic-Arithmetic mean, we have that \[ RHS = n\sqrt {\frac {(x_{1}^{1/2})^{2} + (x_{2}^{1/2})^{2} + ... + (x_{n}^{1/2})^{2}}{n}}\geq n\left[\frac {x_{1}^{1/2} + x_{2}^{1/2} + ... + x_{n}^{1/2}}{n}\right] = x_{1}^{1/2} + x_{2}^{1/2} + ... + x_{n}^{1/2}\] Since $ x_{1},x_{2},...,x_{n}\geq 1$, we have that $ x_{1}^{1/2} = x_{1}^{1/2}$, $ x_{2}^{1/2}\geq x_{2}^{1/3}$, $ x_{3}^{1/2}\geq x_{3}^{1/4}$, ..., $ x_{n}^{1/2}\geq x_{n}^{1/n+1}$ with equalities if $ x_{2},x_{3},...,x_{n} = 1$. Therefore, for the equality $ x_{1}^{\frac {1}{2}} + ... + x_{n}^{\frac {1}{n + 1}} = n\sqrt {\frac {x_{1} + ... + x_{n}}{n}}$ to hold, we need $ x_{2},x_{3},...,x_{n} = 1$. Also, since we are counting on the quadratc mean holding equality, we also need $ x_{1} = 1$. Taking this into account, we see that $ x_{n + 1} = 1$. Then $ x_{1},...,x_{n + 1} = 1$ is the only possible solution.

It would be nice if people who writes the problem, at least they know the problem. I was trying to recopilate problems for a training and now I don´t know the exact problem. Thank you