Find the extremum of $$\frac{ax+by+c}{\sqrt{1+x^2+y^2}}$$ where $a^2+b^2+c^2>0$.
I managed to find critical point - ($\frac{a}{c}$,$\frac{b}{c}$) if $c \neq0$, and no points otherwise. Value of function at this point looks great: $\sqrt{a^2+b^2+c^2}$ if $c>0$, and $-\sqrt{a^2+b^2+c^2}$ otherwise. I suppose it's maximum and minimum, but checking it by calculating determinants looks like suicide. Is there easier way?
You can get it by the well know Cauchy inequality, see here for a proof.
Since by Cauchy's inequality we have $$ (a^2+b^2+c^2)(x^2+y^2+1)\geqslant (ax+by+c)^2. $$