Problem B. 4703. (March 2015)
B. 4703. Given that the absolute values of the numbers \(\displaystyle x_1\), \(\displaystyle x_2\), \(\displaystyle x_3\), \(\displaystyle x_4\), \(\displaystyle x_5\), \(\displaystyle x_6\) are at most 1, and their sum is 0, prove that
\(\displaystyle
3\sum_{i=1}^{5} {\sqrt{1x_i^2}} \le \sum_{i=1}^{5} {\sqrt{9{(x_i+x_{i+1})}^2}}\,.
\)
Suggested by K. Williams, Szeged
(6 pont)
Deadline expired on 10 April 2015.
Statistics:
2 students sent a solution.  
4 points:  1 student. 
0 point:  1 student. 
