There are $m$ real numbers $x_i \geq 0$ ($i=1,2,...,m$), $n \geq 2$, $\sum_{i=1}^{m} x_i=S$. Prove that \[ \sum_{i=1}^{m} \sqrt[n]{\frac{x_i}{S-x_i}} \geq 2, \]The equation holds if and only if there are exactly two of $x_i$ are equal(not equal to $0$), and the rest are equal to $0$.
Problem
Source: 2017 Taiwan TST Round 3
Tags: inequalities
13.04.2018 17:20
Looks like KMO 2010 P5, probably can be solved in a similar way.
13.04.2018 18:01
Korea National Olympiad 2010 Problem 5: $ x, y, z $ are positive real numbers such that $ x+y+z=1 $. Prove that
13.04.2018 18:21
ltf0501 wrote: There are $m$ real numbers $x_i \geq 0$ ($i=1,2,...,m$), $n \geq 2$, $\sum_{i=1}^{m} x_i=S$. Prove that \[ \sum_{i=1}^{m} \sqrt[n]{\frac{x_i}{S-x_i}} \geq 2, \]The equation holds if and only if there are exactly two of $x_i$ are equal(not equal to $0$), and the rest are equal to $0$. See: Let $a_1,a_2,\cdots,a_n \in R^+ ,$ and $ \sum_{i = 1}^{n} a_i=\lambda .$ prove that\[\sum_{i = 1}^{n} \sqrt{\frac{a_i}{\lambda-a_i}}> 2 .\]
27.04.2018 16:15
Sketch: WLOG $S=1$. We will divide the problem into two cases. If $x_i \le \frac{1}{2}$ for all $i$, we can just prove the ineq for $n=2$ only (larger $n$ increases LHS), which is KMO 2010. If $x_1 > \frac{1}{2}$, we can use MV-like technique - using $(u,v) \rightarrow (u+v,0)$, where $u+v < \frac{1}{2}$. This reduces the inequality to trivial two-variable inequality. Done