Consider the form $$ Q(x) = 2q(x) = (x_1+x_2)^2 + (x_2+x_3)^2 + \ldots + (x_7+x_1)^2.$$ You have to show that it has Euclidean minimum $\frac74$ attained at $X_1 = x_1+x_2 = \frac12$, ..., $X_7 = x_7 + x_1 = \frac12$, but unfortunately not over the lattice ${\mathbb Z}^7$, where it would be trivial, but over a lattice of index $2$ defined by the condition $y_1 + y_2 + \ldots + y_7 \equiv 0 \bmod 2$. It is clear that $(\frac12, \ldots, \frac12)$ has Euclidean minimum $\frac 74$, and it remains to show that all other points in the fundamental domain of the lattice have a minimum at most $\frac74$.
This is not difficult to see: assume you have the point $(\frac12 + \delta, \frac12 + \varepsilon, ...)$ for small $\delta, \varepsilon \ge 0$. Subtracting the point $(1,1,0,\ldots, 0)$ you will get a point with coordinates $(-\frac12 + \delta, -\frac12 + \varepsilon, ...)$. Repeating this procedure you will eventually reach a point with $|X_2|, \ldots, |X_7| \le \frac12$. If $|X_1| \le\frac12$, we are done. If not, you can make $|X_1| < \frac12$ at the cost of making another coordinate $> \frac12$ in absolute values. If you think about this for a minute you will see that we can always reach a point of the form $$ X = \Big(\frac12 + \delta_1, \frac12 - \delta_2, . . . , \frac12 - \delta_7\Big)$$ with $0 \le \delta_j \le \frac12$ and $\delta_1 \le \delta_i$ for all $i > 1$. It remains to show that $q(X) \le \frac74$, which is equivalent to $$ \delta_1 - \delta_2 - \ldots - \delta_7 + \delta_1^2 + \ldots + \delta_7^2 \le 0. $$ Does this inequality hold?