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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.

3 Answers 3

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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 $-\frac{1}{2}\sqrt{a^2+b^2}$ .

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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.

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    Happy New Year Joe! (+1) for the nice succinct solution. -Mark – 2017-01-25
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    @Dr.MV: Many happy returns! – 2017-01-25
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    @copper.hat could you please little bit more. How inner problem is straight forward? – 2017-02-01
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    @Toeplitz: I don't want to eclipse Mark's answer. The inner problem is the minimisation of a linear functional ($(x,y) \mapsto ax-by$) over a convex set (the disk), so $\min_{x^2+y^2= r^2} ax-by = r \sqrt{a^2+b^2}$ (Cauchy Schwartz). – 2017-02-01
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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!