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Given

$$M = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2: x^2+y^2 \leq 1\right\} $$

How can one prove that

$$\min \left\{ x+y:\begin{pmatrix} x \\ y \end{pmatrix} \in M \right\}$$

exists and calculate that?

I wasn't able to find anything on math stackexchange regarding this. To calculate the minimum, can't one just use parametrization of $\partial M$ which would lead to $0$?

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  • $\begingroup$ $x=-1,y=0$ is in $M$ so the minimum is at least $-1$. It might help to put $x=r\sin x, y=r\cos x$. Also the ${x \choose y} $ is usually used as binomial coefficient, might be better to use $(x,y) \in M$ $\endgroup$
    – kingW3
    Feb 6, 2020 at 2:05

2 Answers 2

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The easiest way to solve this problem is to use geometry. On a plane, $M$ is a circle of radius $1$ with the center in $(0, 0)$. $x+y=const$ are the straight lines at $135°$; the 'lower' is the line, the smaller is $x+y$.

So you need to find the point where the 'lowest' line that intersects the circle of radius $1$ with the center in $(0, 0)$. This line will be tangent to the circle.

This point is $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.

Does it make any sense at all? I can add a drawing, but it won't be a very good drawing :)

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You can show the existence of the minimum and calculate it using Cauchy-Schwarz inequality:

$$|1\cdot x + 1\cdot y| \leq \sqrt{2}\sqrt{x^2+y^2}\leq\sqrt 2$$

Hence,

$$-\sqrt{2} \leq 1\cdot x + 1\cdot y$$

and equality is reached for $x=y=-\frac{\sqrt 2}{2}$.

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