Proving inequality relationAn information theory inequality which relates to Shannon EntropyInequality...
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Proving inequality relation
An information theory inequality which relates to Shannon EntropyInequality holding for complex numbers in the unit diskSearching for tighter boundsQuestion on a proof of Hilbert's Inequality using Cauchy SchwarzAn application of Cauchy-Schwarz inequalityInequality relation between $l^2$ and $l^infty$ metrics$sum_{i<j}x_i x_j=1$ implies $sum_{ine k}x_i<sqrt{2}$ for some index $k$?Prove Cauchy-Schwarz inequality for two n-dimensional vectorsUsing Jensen's inequality to prove another inequality?Show that $max_{1leq ileq n}|X_i-bar{X}| < frac{(n-1)S}{sqrt{n}}$
$begingroup$
I would like to get some help with the next problem:
I'm trying to prove that $$sum_{i = 1}^n (x_i - y_i)^2 le sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2;;;;;(1).$$
I tried this:
$$ sum_{i = 1}^n (x_i - y_i)^2 =$$
$$= sum_{i = 1}^n [x_i^2 -2x_iy_i + y_i^2 + (-2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i) - (-2x_iz_i + z_i^2) - (z_i^2 - 2z_iy_i)] =$$
$$= sum_{i = 1}^n [(x_i^2 - 2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i + y_i^2) - 2z_i^2 -2x_iy_i + 2x_iz_i + 2z_iy_i] =$$
$$= sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2 - 2sum_{i = 1}^n z_i^2 - 2sum_{i = 1}^n (x_iy_i - x_iz_i - z_iy_i).$$
I wanted to get an expression from which it would be easier to conclude that $sum_{i = 1}^n (x_i - y_i)^2$ is smaller than the right side of the inequality $(1)$, but the last expression isn't useful.
Please, could you give me some advice about what should i do to prove this inequality?
EDIT: I apologise, because i made a huge mistake. It wasn't explicitly stated, so i oversaw it: I need to prove a different inequality:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} le sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
EDIT 2: I think i solved it. Using Minkowski inequality, i got:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} = sqrt{sum_{i = 1}^n |x_i + (-y_i)|^2} = sqrt{sum_{i = 1}^n |x_i - z_i + z_i + (-y_i)|^2} le sqrt{sum_{i = 1}^n |x_i - z_i|^2} + sqrt{sum_{i = 1}^n |z_i - y_i|^2} = sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
inequality
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
$endgroup$
add a comment |
$begingroup$
I would like to get some help with the next problem:
I'm trying to prove that $$sum_{i = 1}^n (x_i - y_i)^2 le sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2;;;;;(1).$$
I tried this:
$$ sum_{i = 1}^n (x_i - y_i)^2 =$$
$$= sum_{i = 1}^n [x_i^2 -2x_iy_i + y_i^2 + (-2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i) - (-2x_iz_i + z_i^2) - (z_i^2 - 2z_iy_i)] =$$
$$= sum_{i = 1}^n [(x_i^2 - 2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i + y_i^2) - 2z_i^2 -2x_iy_i + 2x_iz_i + 2z_iy_i] =$$
$$= sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2 - 2sum_{i = 1}^n z_i^2 - 2sum_{i = 1}^n (x_iy_i - x_iz_i - z_iy_i).$$
I wanted to get an expression from which it would be easier to conclude that $sum_{i = 1}^n (x_i - y_i)^2$ is smaller than the right side of the inequality $(1)$, but the last expression isn't useful.
Please, could you give me some advice about what should i do to prove this inequality?
EDIT: I apologise, because i made a huge mistake. It wasn't explicitly stated, so i oversaw it: I need to prove a different inequality:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} le sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
EDIT 2: I think i solved it. Using Minkowski inequality, i got:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} = sqrt{sum_{i = 1}^n |x_i + (-y_i)|^2} = sqrt{sum_{i = 1}^n |x_i - z_i + z_i + (-y_i)|^2} le sqrt{sum_{i = 1}^n |x_i - z_i|^2} + sqrt{sum_{i = 1}^n |z_i - y_i|^2} = sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
inequality
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
$endgroup$
1
$begingroup$
This seems equivalent to saying 'the length $XY$ squared is no greater than the sum of squares of distances $XZ$ and $ZY$' where X, Y, Z are points in $n$-dimensional Euclidean space, given by their Cartesian coordinates. But that is false: suppose Z is a midpoint of XY, then $|XY|^2 = 4times (|XY|/2)^2 > 2times (|XY|/2)^2 = |XZ|^2+|ZY|^2.$
$endgroup$
– CiaPan
Mar 19 at 16:14
$begingroup$
@CiaPan : Well, i'm trying to prove that function $d(x, y) = sum_{i = 1}^n (x_i - y_i)^2$ is metric. In the book they say it is, but don't provide the proof. Maybe they made a mistake or typo.
$endgroup$
– SlowLearner
Mar 19 at 16:23
add a comment |
$begingroup$
I would like to get some help with the next problem:
I'm trying to prove that $$sum_{i = 1}^n (x_i - y_i)^2 le sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2;;;;;(1).$$
I tried this:
$$ sum_{i = 1}^n (x_i - y_i)^2 =$$
$$= sum_{i = 1}^n [x_i^2 -2x_iy_i + y_i^2 + (-2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i) - (-2x_iz_i + z_i^2) - (z_i^2 - 2z_iy_i)] =$$
$$= sum_{i = 1}^n [(x_i^2 - 2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i + y_i^2) - 2z_i^2 -2x_iy_i + 2x_iz_i + 2z_iy_i] =$$
$$= sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2 - 2sum_{i = 1}^n z_i^2 - 2sum_{i = 1}^n (x_iy_i - x_iz_i - z_iy_i).$$
I wanted to get an expression from which it would be easier to conclude that $sum_{i = 1}^n (x_i - y_i)^2$ is smaller than the right side of the inequality $(1)$, but the last expression isn't useful.
Please, could you give me some advice about what should i do to prove this inequality?
EDIT: I apologise, because i made a huge mistake. It wasn't explicitly stated, so i oversaw it: I need to prove a different inequality:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} le sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
EDIT 2: I think i solved it. Using Minkowski inequality, i got:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} = sqrt{sum_{i = 1}^n |x_i + (-y_i)|^2} = sqrt{sum_{i = 1}^n |x_i - z_i + z_i + (-y_i)|^2} le sqrt{sum_{i = 1}^n |x_i - z_i|^2} + sqrt{sum_{i = 1}^n |z_i - y_i|^2} = sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
inequality
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
$endgroup$
I would like to get some help with the next problem:
I'm trying to prove that $$sum_{i = 1}^n (x_i - y_i)^2 le sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2;;;;;(1).$$
I tried this:
$$ sum_{i = 1}^n (x_i - y_i)^2 =$$
$$= sum_{i = 1}^n [x_i^2 -2x_iy_i + y_i^2 + (-2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i) - (-2x_iz_i + z_i^2) - (z_i^2 - 2z_iy_i)] =$$
$$= sum_{i = 1}^n [(x_i^2 - 2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i + y_i^2) - 2z_i^2 -2x_iy_i + 2x_iz_i + 2z_iy_i] =$$
$$= sum_{i = 1}^n (x_i - z_i)^2 + sum_{i = 1}^n (z_i - y_i)^2 - 2sum_{i = 1}^n z_i^2 - 2sum_{i = 1}^n (x_iy_i - x_iz_i - z_iy_i).$$
I wanted to get an expression from which it would be easier to conclude that $sum_{i = 1}^n (x_i - y_i)^2$ is smaller than the right side of the inequality $(1)$, but the last expression isn't useful.
Please, could you give me some advice about what should i do to prove this inequality?
EDIT: I apologise, because i made a huge mistake. It wasn't explicitly stated, so i oversaw it: I need to prove a different inequality:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} le sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
EDIT 2: I think i solved it. Using Minkowski inequality, i got:
$$sqrt{sum_{i = 1}^n (x_i - y_i)^2} = sqrt{sum_{i = 1}^n |x_i + (-y_i)|^2} = sqrt{sum_{i = 1}^n |x_i - z_i + z_i + (-y_i)|^2} le sqrt{sum_{i = 1}^n |x_i - z_i|^2} + sqrt{sum_{i = 1}^n |z_i - y_i|^2} = sqrt{sum_{i = 1}^n (x_i - z_i)^2} + sqrt{sum_{i = 1}^n (z_i - y_i)^2}.$$
inequality
inequality
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
edited Mar 19 at 17:19
SlowLearner
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
asked Mar 19 at 16:00
SlowLearnerSlowLearner
18513
18513
1
$begingroup$
This seems equivalent to saying 'the length $XY$ squared is no greater than the sum of squares of distances $XZ$ and $ZY$' where X, Y, Z are points in $n$-dimensional Euclidean space, given by their Cartesian coordinates. But that is false: suppose Z is a midpoint of XY, then $|XY|^2 = 4times (|XY|/2)^2 > 2times (|XY|/2)^2 = |XZ|^2+|ZY|^2.$
$endgroup$
– CiaPan
Mar 19 at 16:14
$begingroup$
@CiaPan : Well, i'm trying to prove that function $d(x, y) = sum_{i = 1}^n (x_i - y_i)^2$ is metric. In the book they say it is, but don't provide the proof. Maybe they made a mistake or typo.
$endgroup$
– SlowLearner
Mar 19 at 16:23
add a comment |
1
$begingroup$
This seems equivalent to saying 'the length $XY$ squared is no greater than the sum of squares of distances $XZ$ and $ZY$' where X, Y, Z are points in $n$-dimensional Euclidean space, given by their Cartesian coordinates. But that is false: suppose Z is a midpoint of XY, then $|XY|^2 = 4times (|XY|/2)^2 > 2times (|XY|/2)^2 = |XZ|^2+|ZY|^2.$
$endgroup$
– CiaPan
Mar 19 at 16:14
$begingroup$
@CiaPan : Well, i'm trying to prove that function $d(x, y) = sum_{i = 1}^n (x_i - y_i)^2$ is metric. In the book they say it is, but don't provide the proof. Maybe they made a mistake or typo.
$endgroup$
– SlowLearner
Mar 19 at 16:23
1
1
$begingroup$
This seems equivalent to saying 'the length $XY$ squared is no greater than the sum of squares of distances $XZ$ and $ZY$' where X, Y, Z are points in $n$-dimensional Euclidean space, given by their Cartesian coordinates. But that is false: suppose Z is a midpoint of XY, then $|XY|^2 = 4times (|XY|/2)^2 > 2times (|XY|/2)^2 = |XZ|^2+|ZY|^2.$
$endgroup$
– CiaPan
Mar 19 at 16:14
$begingroup$
This seems equivalent to saying 'the length $XY$ squared is no greater than the sum of squares of distances $XZ$ and $ZY$' where X, Y, Z are points in $n$-dimensional Euclidean space, given by their Cartesian coordinates. But that is false: suppose Z is a midpoint of XY, then $|XY|^2 = 4times (|XY|/2)^2 > 2times (|XY|/2)^2 = |XZ|^2+|ZY|^2.$
$endgroup$
– CiaPan
Mar 19 at 16:14
$begingroup$
@CiaPan : Well, i'm trying to prove that function $d(x, y) = sum_{i = 1}^n (x_i - y_i)^2$ is metric. In the book they say it is, but don't provide the proof. Maybe they made a mistake or typo.
$endgroup$
– SlowLearner
Mar 19 at 16:23
$begingroup$
@CiaPan : Well, i'm trying to prove that function $d(x, y) = sum_{i = 1}^n (x_i - y_i)^2$ is metric. In the book they say it is, but don't provide the proof. Maybe they made a mistake or typo.
$endgroup$
– SlowLearner
Mar 19 at 16:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is enought to prove for $n=1$: $$(x-y)^2leq (x-z)^2+(z-y)^2$$
It is the same as $$0leq 2z^2-2xz-2yz+2xy$$
and this is $$(z-x)(z-y)geq 0$$ which is not true if, say $x=1,z=2$ and $y=3$.
So you have forgotten something.
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
$endgroup$
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
1
$begingroup$
Never mind, just want to help
$endgroup$
– Maria Mazur
Mar 19 at 16:39
$begingroup$
I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
$endgroup$
– SlowLearner
Mar 19 at 16:41
|
show 2 more comments
Your Answer
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It is enought to prove for $n=1$: $$(x-y)^2leq (x-z)^2+(z-y)^2$$
It is the same as $$0leq 2z^2-2xz-2yz+2xy$$
and this is $$(z-x)(z-y)geq 0$$ which is not true if, say $x=1,z=2$ and $y=3$.
So you have forgotten something.
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
$endgroup$
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
1
$begingroup$
Never mind, just want to help
$endgroup$
– Maria Mazur
Mar 19 at 16:39
$begingroup$
I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
$endgroup$
– SlowLearner
Mar 19 at 16:41
|
show 2 more comments
$begingroup$
It is enought to prove for $n=1$: $$(x-y)^2leq (x-z)^2+(z-y)^2$$
It is the same as $$0leq 2z^2-2xz-2yz+2xy$$
and this is $$(z-x)(z-y)geq 0$$ which is not true if, say $x=1,z=2$ and $y=3$.
So you have forgotten something.
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
$endgroup$
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
1
$begingroup$
Never mind, just want to help
$endgroup$
– Maria Mazur
Mar 19 at 16:39
$begingroup$
I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
$endgroup$
– SlowLearner
Mar 19 at 16:41
|
show 2 more comments
$begingroup$
It is enought to prove for $n=1$: $$(x-y)^2leq (x-z)^2+(z-y)^2$$
It is the same as $$0leq 2z^2-2xz-2yz+2xy$$
and this is $$(z-x)(z-y)geq 0$$ which is not true if, say $x=1,z=2$ and $y=3$.
So you have forgotten something.
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
$endgroup$
It is enought to prove for $n=1$: $$(x-y)^2leq (x-z)^2+(z-y)^2$$
It is the same as $$0leq 2z^2-2xz-2yz+2xy$$
and this is $$(z-x)(z-y)geq 0$$ which is not true if, say $x=1,z=2$ and $y=3$.
So you have forgotten something.
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
edited Mar 19 at 16:19
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
answered Mar 19 at 16:04
Maria MazurMaria Mazur
49.7k1361124
49.7k1361124
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
1
$begingroup$
Never mind, just want to help
$endgroup$
– Maria Mazur
Mar 19 at 16:39
$begingroup$
I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
$endgroup$
– SlowLearner
Mar 19 at 16:41
|
show 2 more comments
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
1
$begingroup$
Never mind, just want to help
$endgroup$
– Maria Mazur
Mar 19 at 16:39
$begingroup$
I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
$endgroup$
– SlowLearner
Mar 19 at 16:41
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
I didn't forget anything, which is even more confusing.
$endgroup$
– SlowLearner
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
Which book is this?
$endgroup$
– Maria Mazur
Mar 19 at 16:24
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
$begingroup$
I don't know. I only have photos of the pages of the first chapter. I took the book from the library, but i didn't want to copy it whole, because of possible damaging, so i just made photos of the pages i need. :) I guess i don't care about copyrights.
$endgroup$
– SlowLearner
Mar 19 at 16:37
1
1
$begingroup$
Never mind, just want to help
$endgroup$
– Maria Mazur
Mar 19 at 16:39
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Never mind, just want to help
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– Maria Mazur
Mar 19 at 16:39
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I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
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– SlowLearner
Mar 19 at 16:41
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I'm sorry, but i didn't remember authors or the exact title. Anyway, it's about analysis and it's in Serbian.
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– SlowLearner
Mar 19 at 16:41
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This seems equivalent to saying 'the length $XY$ squared is no greater than the sum of squares of distances $XZ$ and $ZY$' where X, Y, Z are points in $n$-dimensional Euclidean space, given by their Cartesian coordinates. But that is false: suppose Z is a midpoint of XY, then $|XY|^2 = 4times (|XY|/2)^2 > 2times (|XY|/2)^2 = |XZ|^2+|ZY|^2.$
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– CiaPan
Mar 19 at 16:14
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@CiaPan : Well, i'm trying to prove that function $d(x, y) = sum_{i = 1}^n (x_i - y_i)^2$ is metric. In the book they say it is, but don't provide the proof. Maybe they made a mistake or typo.
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– SlowLearner
Mar 19 at 16:23