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If $x^2+y^2 \leq 1$, then $|x+y|\leq 1$?
It's probably very easy, but I can't solve it : )

It's a part of bigger problem.

I have to do this one: I know that $x^2+y^2 \leq 1$. I must find a smallest number n which fulfills $|x+y|+|x-y|\leq n$. It's probably $2$, but I don't know how to show that.

I know that $|x| \leq 1$, $|y| \leq 1$. It's easy to show that $|x+y|+|x-y| \leq |x|+|y|+|x|+|y| \leq 4$.
But how to get $2$.

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    $\begingroup$ Have you made any attempts on the problem? Where are you stuck? $\endgroup$ – Harnoor Lal Feb 21 '17 at 19:01
  • $\begingroup$ Sharpest bound would be $|x+y|\leq \sqrt{2}$, 1 is not possible. $\endgroup$ – Shuhao Cao Feb 21 '17 at 19:07
  • $\begingroup$ Please use an informative title to your question, to help others find it later. $\endgroup$ – David G. Stork Feb 21 '17 at 19:09
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    $\begingroup$ "I must find a smallest number n which fulfills |x+y|+|x-y|≤n" Try squaring both sides. $\endgroup$ – Doug M Feb 21 '17 at 19:17
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Take $\;x=y=\cfrac1{\sqrt2}\;$ and get a straighforward counterexample ...

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It is wrong. For example, if $x=y=\frac{\sqrt2}{2}$, then $x^2+y^2=1$ but $|x+y|=\sqrt2>1$.

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No. Restrict your attention to $x,y \geq 0$ for a moment.

The first inequality says that $y \leq \sqrt{1-x^2}$ which is a quarter circle with radius $1$. The second is the area in between the line $y=1-x$ and the two axes.

Does being in a quarter circle imply being below the line?

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  • $\begingroup$ This approach also makes it clear how one should weaken the second inequality to make it always valid. $\endgroup$ – Semiclassical Feb 21 '17 at 19:13
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See graphics for $x^2+y^2=1$ and $\vert x + y \vert=1 $

enter image description here

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This is not true, take $x=\frac{1}{2}$ and $y=\frac{\sqrt{3}}{2}$, then $x^2+y^2=1$ but $x+y =\frac{1+\sqrt{3}}{2} > \frac{1+1}{2} = 1$

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This is not true, try to characterise the set of solutions $(x,y)$ in the plane..

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we are trying to find $|x+y|+|x-y|\le n$. So by the triangle inequality, $$ |2x|=|x+y+x-y|\le |x+y|+|x-y|\le n$$ Since $x^2+y^2\le 1$ $x$ we have $$|x|\le x^2\le x^2+y^2\le 1$$ This implies that $$2|x|=|2x|\le 1+1=2$$ Therefore $n=2$

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  • $\begingroup$ @Joe The rest of what you were trying to do is superfluous and false as the earlier answers can easily show. $\endgroup$ – Sentinel135 Feb 21 '17 at 19:57
  • $\begingroup$ for $ x=\sqrt{2}/2 $ and $ y=\sqrt{2}/2 $ we have $|x+y|+|x-y|\le \sqrt{2}$, but if the inequality must be fulfilled for all points of the unit circle then really $n=2$ is the solution.. $\endgroup$ – Widawensen Feb 21 '17 at 20:04
  • $\begingroup$ If we were trying to find the smallest n for which the inequality $|x+y|+|x-y|\le n$ holds then the smallest $n$ would be 0, since $x=y=0$ is a solution for $x^2+y^2\le 1$. I assumed he meant: Find the smallest $n$ such that $|x+y|+|x-y|\le n$ $\forall x,y\in \mathbb R$ where $x^2+y^2\le 1$. $\endgroup$ – Sentinel135 Feb 21 '17 at 20:45
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$|x+y| + |x-y| \le n$

Square both sides:

$(x+y)^2 + (x-y)^2 + 2|(x+y)(x-y)| \le n^2\\ 2x^2 + 2y^2 + 2|x^2-y^2| \le n^2$

$|x^2-y^2| = (x^2 + y^2) \cos (2\arctan \frac yx) < (x^2 + y^2)\\ x^2 + y^2 \le 1\\ (|x+y| + |x-y|)^2 \le 4 (x^2 + y^2) \le 4 \le n^2\\ n \ge 2$

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