7. Let $x, y$ be real numbers, and $x^{2}-2 x y+y^{2}-\sqrt{2} x-\sqrt{2} y+6=0, u=x+y$ has the minimum value of $\qquad$ .
7. $3 \sqrt{2}$.
Let $y=u-x$, substituting into the known equation gives $4 x^{2}-4 u x+u^{2}-\sqrt{2} u+6=0, \Delta=16 u^{2}-16\left(u^{2}-\sqrt{2} u+6\right) \geqslant 0$, which yields $u \geqslant 3 \sqrt{2}$. When $x=y=\frac{3 \sqrt{2}}{2}$, $u=3 \sqrt{2}$.