Let $a_1, a_2, ...,a_n$ be positive real numbers such that $$a_1 + a_2 + ... + a_n = a_1^2 + a_2^2 + ... + a_n^2$$Prove that $$\sum_{1\le i<j\le n} a_ia_j(1 - a_ia_j) \ge 0$$Võ Quốc Bá Cẩn.
Problem
Source: 2015 Saudi Arabia IMO TST II p3
Tags: inequalities, algebra, n-variable inequality
24.07.2020 19:25
We may write the inequality as $$\sum_{1 \le i<j \le n} a_ia_j \ge \sum_{1 \le i<j \le n} a_i^2a_j^2.$$From the identities $$\sum_{1 \le i < j \le n} a_ia_j = \frac{1}{2}\left(\left(\sum_{i=1}^n a_i\right)^2 - \sum_{i=1}^n a_i^2\right), \ \sum_{1 \le i < j \le n} a_i^2a_j^2 = \frac{1}{2}\left(\left(\sum_{i=1}^n a_i^2\right)^2 - \sum_{i=1}^n a_i^4\right)$$and the condition $\sum_{i=1}^n a_i = \sum_{i=1}^n a_i^2$, the inequality is equivalent to $$\sum_{i=1} a_i^4 \ge \sum_{i=1} a_i^2.$$By Hölder's inequality, we have $$\left(\sum_{i=1}^n a_i^4\right)\left(\sum_{i=1}^n a_i\right)^2 \ge \left(\sum_{i=1}^n a_i^2\right)^3.$$From $\sum_{i=1}^n a_i = \sum_{i=1}^n a_i^2$, the inequality becomes $$\sum_{i=1} a_i^4 \ge \sum_{i=1} a_i^2$$as we want.
24.07.2020 21:33
NaPrai wrote: We may write the inequality as $$\sum_{1 \le i<j \le n} a_ia_j \ge \sum_{1 \le i<j \le n} a_i^2a_j^2.$$From the identities $$\sum_{1 \le i < j \le n} a_ia_j = \frac{1}{2}\left(\left(\sum_{i=1}^n a_i\right)^2 - \sum_{i=1}^n a_i^2\right), \ \sum_{1 \le i < j \le n} a_i^2a_j^2 = \frac{1}{2}\left(\left(\sum_{i=1}^n a_i^2\right)^2 - \sum_{i=1}^n a_i^4\right)$$and the condition $\sum_{i=1}^n a_i = \sum_{i=1}^n a_i^2$, the inequality is equivalent to $$\sum_{i=1} a_i^4 \ge \sum_{i=1} a_i^2.$$By Hölder's inequality, we have $$\left(\sum_{i=1}^n a_i^4\right)\left(\sum_{i=1}^n a_i\right)^2 \ge \left(\sum_{i=1}^n a_i^2\right)^3.$$From $\sum_{i=1}^n a_i = \sum_{i=1}^n a_i^2$, the inequality becomes $$\sum_{i=1} a_i^4 \ge \sum_{i=1} a_i^2$$as we want. Your solution is awesome! I am kinda new in olympiad inequalities and summation signs really scare me, i would be really thankful if you could share any resource that you used to improve your skills in this topic Best regards Thank you