oVlad wrote: Let $a_1,a_2,a_3,a_4$ be positive real numbers satisfying \[\sum_{i<j}a_ia_j=1.\]Prove that \[\sum_{\text{sym}}\frac{a_1a_2}{1+a_3a_4}\geq\frac{6}{7}.\]* * * We have $6\sqrt{a_1a_2a_3a_4}\leq\sum_{i<j}a_ia_j=1$ and \[ \sum_{\text{sym}}\frac{a_1a_2}{1+a_3a_4}=\sum_{\text{sym}}\frac{a_1^2a_2^2}{a_1a_2+a_1a_2a_3a_4}\geq\frac{\left(\sum_{i<j}a_ia_j\right)^2}{\left(\sum_{i<j}a_ia_j\right)+6a_1a_2a_3a_4}\geq\frac{1^2}{1+6\cdot\left(\frac{1}{6}\right)^2}=\frac{6}{7} \]