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.
 A: Here is an approach that is essentially identical to Mark's answer:
Note that $\inf_{x^2+y^2 \le 1} f(x,y)= \inf_{r \in [0,1]} \inf_{x^2+y^2 = r^2} f(x,y)$. The 'inner' problem $\inf_{x^2+y^2 = r^2} f(x,y)$ is straightforward to solve.
A: It might be easier to work in polar coordinates $(r,\theta)$.  Note that 
$$f(x,y)=\frac{ax-by}{1+x^2+y^2}=\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}=g(r,\theta)$$
Then, we have
$$\begin{align}
\frac{\partial}{\partial r}\left(\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}\right)&=\frac{(1-r^2)}{(1+r^2)^2}\,(a\cos(\theta)-b\sin(\theta)) \tag 1\\\\
\frac{\partial}{\partial \theta}\left(\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}\right)&=-\frac{r}{1+r^2}(a\sin(\theta)+b\cos(\theta))\tag 2
\end{align}$$
Setting $(1)$ to zero, yields $r=1$ or $\tan(\theta)=a/b$.  Setting $(2)$ to zero, yields $r=0$ or $\tan(\theta)=-b/a$.
Hence, the possible local extrema are $r=1$, $\tan(\theta)=-b/a$ or $r=0$ and $\tan(\theta)=a/b$.  
Note that $g(0,\theta)=0$, while $g(1,\arctan(-b/a))=\pm \frac12 \sqrt{a^2+b^2}$.  Therefore, the minimum value of $g(r,\theta)$ is $-\frac12 \sqrt{a^2+b^2}$ which occurs when $r=1$, $\cos(\theta)=-a/\sqrt{a^2+b^2}$, and $\sin(\theta)=b/\sqrt{a^2+b^2}$.  
Transforming to Cartesian coordinates, the minimum of $f(x,y)$ is $-\frac12 \sqrt{a^2+b^2}$, which occurs when $x=-a/\sqrt{a^2+b^2}$ and $y=b/\sqrt{a^2+b^2}$ as conjectured!
A: 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 $-\frac{1}{2}\sqrt{a^2+b^2}$ .
