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.

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