7. Find the minimum value of the function $f(x, y)=\frac{2015(x+y)}{\sqrt{2015 x^{2}+2015 y^{2}}}$ and specify all pairs $(x, y)$ at which it is achieved.
Answer: a) $-\sqrt{4030}$. b) $x=y<0$.

Solution: $f(x, y)= \pm \sqrt{2015} \sqrt{\frac{(x+y)^{2}}{x^{2}+y^{2}}}$. The minimum value is achieved when $\frac{(x+y)^{2}}{x^{2}+y^{2}}$ reaches its maximum, and $x+y<0$. Note that $\frac{(x+y)^{2}}{x^{2}+y^{2}}=1+\frac{2 x y}{x^{2}+y^{2}} \leqslant 2$ and equality is achieved when $x=y$. Therefore, the minimum of the original function is $-\sqrt{2015} \cdot \sqrt{2}$.