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

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 $$0$$?

• $x=-1,y=0$ 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 $(x,y) \in M$ Feb 6, 2020 at 2:05

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 :)

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 $$x=y=-\frac{\sqrt 2}{2}$$.