I tried to prove it like this:
$\sqrt{x-a}^2+\sqrt{y-b}^2=\sqrt{(\sqrt{x-a}^2+\sqrt{y-b}^2)^2}$
from this we can expand all the terms and end up with a messy sum that I suppose results in
$|$(x+y)-(a+b)|
Is there more sophisticated and logical way of proving or should I say deriving this?
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ToTheSpace 2
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Focus on the properties of absolute values. $|a|=a$, where is $\pm a$. – cemsicles Nov 11 '19 at 14:16
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3You cannot; try with $a=b=0$ and $x=1, y=-1$. We have $|x−a|+|y−b|=|x|+|y|=2$ and $|(x+y)−(a+b)|=|(x+y)|=0$. – Mauro ALLEGRANZA Nov 11 '19 at 14:17
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Perfection! @MauroALLEGRANZA . Inequality is possible only. – Sujit Bhattacharyya Nov 11 '19 at 14:18
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1With $u=x-a$ and $v =y-b$ you have $|u| + |v| = |u+v|$ and that holds (for real numbers) if and only if $u$ and $v$ have the same sign (or one of them is zero). – Martin R Nov 11 '19 at 14:20
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I see now. This is only true for inequalities. $|x−a|+|y−b|>=|(x+y)−(a+b)|$ Now I need to prove this. – ToTheSpace 2 Nov 11 '19 at 14:20
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With $u=x-a$ and $v =y-b$ you have the triangle inequality $$ |(x+y) - (a+b)| =|u+v| \le |u| + |v| = |x-a| + |y-b| \, . $$
Equality holds (for real numbers) if and only if $uv \ge 0$, i.e. $x-a$ and $y-b$ have the same sign, or one of them is zero.
For various proofs of the triangle inequality for real numbers, see for example Proof of triangle inequality or Triangle Inequality/Real Numbers.
Martin R
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