Let $m<n$ be two positive integers, let $I$ and $J$ be two index sets such that $|I|=|J|=n$ and $|I\cap J|=m$, and let $u_k$, $k\in I\cup J$ be a collection of vectors in the Euclidean plane such that \[|\sum_{i\in I}u_i|=1=|\sum_{j\in J}u_j|.\] Prove that \[\sum_{k\in I\cup J}|u_k|^2\geq \frac{2}{m+n}\] and find the cases of equality.
Problem
Source: Romania TST 4 2009, Problem 2
Tags: inequalities, vector, geometry proposed, geometry
dgrozev
08.05.2012 12:16
An easy one for Romania TST! We have: (1) $ \sum_{k\in I\cup J}|u_{k}|^{2}= \sum_{k \in I \cap J} |u_k|^2 + \sum_{k \in I \setminus J} |u_k|^2 + \sum_{k \in J \setminus I} |u_k|^2 \geq $ $\geq \frac{1}{m}|\sum_{k \in I \cap J} u_k|^2 + \frac{1}{n-m}|\sum_{k \in I \setminus J} u_k|^2 + \frac{1}{n-m}|\sum_{k \in J \setminus I} u_k|^2$ Denote $x= |\sum_{k \in I \cap J} u_k| $. Then (1) can be continued as follows: (2) $\cdots \geq \frac{x^2}{m} + \frac{2(1-x)^2}{n-m}$ RHS of (2) attains its min when $x_{\text{min}} =\frac{2m}{n+m}$ and an easy calculation shows that: $\frac{x_{\text{min}}^2}{m} + \frac{2(1-x_{\text{min}})^2}{n-m} = \frac{2}{n+m}$ So: $ \sum_{k\in I\cup J}|u_{k}|^{2} \geq \frac{2}{n+m} $ (1) and (2) become equalities when all vectors are collinear and with same directions and: $u_k,\, k\in I \cap J$ are equal, $u_k,\, k\in I-J $ are equal $u_k,\, k\in J-I$ are equal. $|\sum_{k \in I \cap J} u_k | = \frac{2m}{n+m}$ One can easily calculate the exact values of $u_k$
Drytime
08.05.2012 21:48
Not easy, but very easy, dgrozev! Actually, the problems in romanian TSTs look harder than they actually are.