(p-q)^2 + (q-r)^2 + (r-p)^2 \ge 0 \Rightarrow \sum p^2 \ge \sum pq \\ \\ \Rightarrow \sum p^2 +2\sum pq \ge 3\sum pq \Rightarrow \left(\sum p)^2 \ge 3 \sum pq
2 Answers
Hari Shankar
·2010-03-29 03:43:10
Hari Shankar
·2010-03-29 07:46:31
There, the starting point is \sum_{i \ne j} \left(\alpha_i - \alpha_j)^2 \ge 0 ; 1 \le i,j \le 5
Verify that each \alpha_i^2 term appears 4 times i.e. the coeff is 4. The coeff of \alpha_i \alpha_j is of course 2.
Can you work it out now