Infimum of two variable function on the closed unit disc I am trying to compute
$$\inf_{x,y} \frac{ax-by}{1+x^2+y^2}$$
subject to the constraint $x^2+y^2 \leq 1$. Here $a,b$ are any two fixed, real numbers. I am having trouble computing this using standard derivative techniques, and Wolfram alpha is unable to recognize $a,b$ as constants. I have plugged in $\pi$ and $e$ as constants into WA and gotten a result, but this obviously is not sufficient when trying to achieve a general result.
I suspect the answer is (assuming $a,b>0$, for example), that this infimum occurs at $x= \frac{-a}{\sqrt{a^2+b^2}}$ and $y= \frac{b}{\sqrt{a^2+b^2}}$, but I cannot rigorously show it, or get WA to give me a general answer.

By using the Cauchy-Schwarz inequality, one has
$$ |f(x,y)|\le\frac{\sqrt{a^2+b^2}\sqrt{x^2+y^2}}{1+x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}+\frac{1}{\sqrt{x^2+y^2}}} $$
and "=" holds if and only if $\frac{a}{b}=\frac{x}{-y}$.
Using
$$ a^2+b^2\ge2ab$$
("=" holds if and only if $a=b$) one has
$$ |f(x,y)|\le\frac{\sqrt{a^2+b^2}\sqrt{x^2+y^2}}{1+x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}+\frac{1}{\sqrt{x^2+y^2}}}\le\frac{1}{2}\sqrt{a^2+b^2}. $$
Note that "=" holds if and only if
$$ \frac{a}{b}=\frac{x}{-y}, x^2+y^2=1.$$
Solving this, one has 
$$ x=\mp\frac{a}{\sqrt{a^2+b^2}}, y=\pm\frac{b}{\sqrt{a^2+b^2}}.$$
Thus
$$ -\frac{1}{2}\sqrt{a^2+b^2}\le f(x)\le \frac{1}{2}\sqrt{a^2+b^2} $$
and hence when $x=-\frac{a}{\sqrt{a^2+b^2}}, y=\frac{b}{\sqrt{a^2+b^2}}$ the minimum of $f(x,y)$ has $-\frac{1}{2}\sqrt{a^2+b^2}$ .
