Triangle Inequality lower and upper bounds?Proving AM ≥ GM in 3 variables using the Cauchy-Schwarz inequalityeuler triangle inequality proof without wordsInequality problem with two modulusConfusion regarding derivation of triangle inequality from Schwarz' inequalityProve triangle inequality using the properties of absolute valueTriangle inequality for specific vector distance functionProof of Triangle Inequality Wikipedia Proof ExplanationA third triangle inequality?Triangle inequality and Minkowski inequalityKantorovich inequality and Cauchy-Schwarz inequality
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Triangle Inequality lower and upper bounds?
Proving AM ≥ GM in 3 variables using the Cauchy-Schwarz inequalityeuler triangle inequality proof without wordsInequality problem with two modulusConfusion regarding derivation of triangle inequality from Schwarz' inequalityProve triangle inequality using the properties of absolute valueTriangle inequality for specific vector distance functionProof of Triangle Inequality Wikipedia Proof ExplanationA third triangle inequality?Triangle inequality and Minkowski inequalityKantorovich inequality and Cauchy-Schwarz inequality
$begingroup$
I was solving a question for linear algebra today and had a question regarding triangle inequalities. The question is:
If $Vert vecv Vert = 5$ and $Vert vecw Vert = 3$, what are the smallest and largest values of $Vert vecv - vecw Vert$?
The way that I solved it is to draw two circles with the origin as their centers and each having radius $5$ and $3$. In this case, $vecv$ and $vecw$ would be each circle's radius. It's not hard to see that the smallest value we can obtain is $2$ and the largest is $8$.
However, I attempted to try solving this question using the triangle inequality. I noticed that the way to solve it is $| Vert vecv Vert - Vert vecw Vert | le Vert vecv - vecw Vert le Vert vecv Vert + Vert vecw Vert$.
The first part I recognize as being the reverse triangle inequality, but how was the second part derived? In my head the original form of the inequality is:
$$Vert vecv + vecw Vert le Vert vecv Vert + Vert vecw Vert$$
with addition not subtraction.
Thank you.
linear-algebra inequality
asked Mar 30 at 4:55
SeankalaSeankala
26011
$endgroup$
add a comment |
$begingroup$
I was solving a question for linear algebra today and had a question regarding triangle inequalities. The question is:
If $Vert vecv Vert = 5$ and $Vert vecw Vert = 3$, what are the smallest and largest values of $Vert vecv - vecw Vert$?
The way that I solved it is to draw two circles with the origin as their centers and each having radius $5$ and $3$. In this case, $vecv$ and $vecw$ would be each circle's radius. It's not hard to see that the smallest value we can obtain is $2$ and the largest is $8$.
However, I attempted to try solving this question using the triangle inequality. I noticed that the way to solve it is $| Vert vecv Vert - Vert vecw Vert | le Vert vecv - vecw Vert le Vert vecv Vert + Vert vecw Vert$.
The first part I recognize as being the reverse triangle inequality, but how was the second part derived? In my head the original form of the inequality is:
$$Vert vecv + vecw Vert le Vert vecv Vert + Vert vecw Vert$$
with addition not subtraction.
Thank you.
linear-algebra inequality
asked Mar 30 at 4:55
SeankalaSeankala
26011
$endgroup$
add a comment |
$begingroup$
I was solving a question for linear algebra today and had a question regarding triangle inequalities. The question is:
If $Vert vecv Vert = 5$ and $Vert vecw Vert = 3$, what are the smallest and largest values of $Vert vecv - vecw Vert$?
The way that I solved it is to draw two circles with the origin as their centers and each having radius $5$ and $3$. In this case, $vecv$ and $vecw$ would be each circle's radius. It's not hard to see that the smallest value we can obtain is $2$ and the largest is $8$.
However, I attempted to try solving this question using the triangle inequality. I noticed that the way to solve it is $| Vert vecv Vert - Vert vecw Vert | le Vert vecv - vecw Vert le Vert vecv Vert + Vert vecw Vert$.
The first part I recognize as being the reverse triangle inequality, but how was the second part derived? In my head the original form of the inequality is:
$$Vert vecv + vecw Vert le Vert vecv Vert + Vert vecw Vert$$
with addition not subtraction.
Thank you.
linear-algebra inequality
asked Mar 30 at 4:55
SeankalaSeankala
26011
$endgroup$
I was solving a question for linear algebra today and had a question regarding triangle inequalities. The question is:
If $Vert vecv Vert = 5$ and $Vert vecw Vert = 3$, what are the smallest and largest values of $Vert vecv - vecw Vert$?
The way that I solved it is to draw two circles with the origin as their centers and each having radius $5$ and $3$. In this case, $vecv$ and $vecw$ would be each circle's radius. It's not hard to see that the smallest value we can obtain is $2$ and the largest is $8$.
However, I attempted to try solving this question using the triangle inequality. I noticed that the way to solve it is $| Vert vecv Vert - Vert vecw Vert | le Vert vecv - vecw Vert le Vert vecv Vert + Vert vecw Vert$.
The first part I recognize as being the reverse triangle inequality, but how was the second part derived? In my head the original form of the inequality is:
$$Vert vecv + vecw Vert le Vert vecv Vert + Vert vecw Vert$$
with addition not subtraction.
Thank you.
linear-algebra inequality
linear-algebra inequality
asked Mar 30 at 4:55
SeankalaSeankala
26011
asked Mar 30 at 4:55
SeankalaSeankala
26011
asked Mar 30 at 4:55
SeankalaSeankala
26011
asked Mar 30 at 4:55
SeankalaSeankala
26011
asked Mar 30 at 4:55
SeankalaSeankala
26011
26011
add a comment |
add a comment |
1 Answer
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$begingroup$
The minimum is $2$ and maximum is $8$. Hint: take $v =cw$ where $c$ is a scalar to see that these values are actually possible.
$|v-w|=|v+(-w)|leq |v|+|w|$ because $|-w|=|w|$.
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
The minimum is $2$ and maximum is $8$. Hint: take $v =cw$ where $c$ is a scalar to see that these values are actually possible.
$|v-w|=|v+(-w)|leq |v|+|w|$ because $|-w|=|w|$.
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
add a comment |
$begingroup$
The minimum is $2$ and maximum is $8$. Hint: take $v =cw$ where $c$ is a scalar to see that these values are actually possible.
$|v-w|=|v+(-w)|leq |v|+|w|$ because $|-w|=|w|$.
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
add a comment |
$begingroup$
The minimum is $2$ and maximum is $8$. Hint: take $v =cw$ where $c$ is a scalar to see that these values are actually possible.
$|v-w|=|v+(-w)|leq |v|+|w|$ because $|-w|=|w|$.
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
The minimum is $2$ and maximum is $8$. Hint: take $v =cw$ where $c$ is a scalar to see that these values are actually possible.
$|v-w|=|v+(-w)|leq |v|+|w|$ because $|-w|=|w|$.
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
answered Mar 30 at 4:57
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
73.3k53170
add a comment |
add a comment |
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