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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?

Some
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1 Answers1

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You can get it by the well know Cauchy inequality, see here for a proof.

Since by Cauchy's inequality we have $$ (a^2+b^2+c^2)(x^2+y^2+1)\geqslant (ax+by+c)^2. $$

Nirvanacs
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