More fun in 2015, Part 30

Since $q(x_1,\ldots,x_{2015}) \; = \; 19\left[ \left(\sum_{i=1}^{2015}x_i\right)^2 - \sum_{i=1}^{2015}x_i^2\right] \; = \; 19\left[\left(\sum_{i=1}^{2015}x_i\right)^2 - 91\right]$ it is clear that $q \ge -19\times91 = -1729$ . It remains to show that we can achieve this value. Since $x_1 = \sqrt{\tfrac{91}{2}}\,, \quad x_2 = -\sqrt{\tfrac{91}{2}}\,, \quad x_3 = x_4 = \cdots = x_{2015} = 0$ gives $\sum_{i=1}^{2015} x_i = 0 \qquad \qquad \sum_{i=1}^{2015} x_i^2 = 91$ the minimum value of $-1729$ can be achieved.