Proof of the 'second' triangle inequality
$begingroup$
I am trying to prove the 'second' triangle inequality:
$$||x|-|y|| leq |x-y|$$
My attempt:
$$----------------$$
Proof:
$|x-y|^2 = (x-y)^2 = x^2 - 2xy + y^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$
Therefore $rightarrow |x-y| geq ||x|-|y||$
$$----------------$$
My questions are: Is this an acceptable proof, and are there alternative proofs that are more efficient?
proof-verification alternative-proof
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
$endgroup$
add a comment |
$begingroup$
I am trying to prove the 'second' triangle inequality:
$$||x|-|y|| leq |x-y|$$
My attempt:
$$----------------$$
Proof:
$|x-y|^2 = (x-y)^2 = x^2 - 2xy + y^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$
Therefore $rightarrow |x-y| geq ||x|-|y||$
$$----------------$$
My questions are: Is this an acceptable proof, and are there alternative proofs that are more efficient?
proof-verification alternative-proof
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
$endgroup$
$begingroup$
it seems good to me.
$endgroup$
– mookid
Mar 25 '14 at 2:15
add a comment |
$begingroup$
I am trying to prove the 'second' triangle inequality:
$$||x|-|y|| leq |x-y|$$
My attempt:
$$----------------$$
Proof:
$|x-y|^2 = (x-y)^2 = x^2 - 2xy + y^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$
Therefore $rightarrow |x-y| geq ||x|-|y||$
$$----------------$$
My questions are: Is this an acceptable proof, and are there alternative proofs that are more efficient?
proof-verification alternative-proof
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
$endgroup$
I am trying to prove the 'second' triangle inequality:
$$||x|-|y|| leq |x-y|$$
My attempt:
$$----------------$$
Proof:
$|x-y|^2 = (x-y)^2 = x^2 - 2xy + y^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$
Therefore $rightarrow |x-y| geq ||x|-|y||$
$$----------------$$
My questions are: Is this an acceptable proof, and are there alternative proofs that are more efficient?
proof-verification alternative-proof
proof-verification alternative-proof
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
asked Mar 25 '14 at 2:12
Display NameDisplay Name
803933
803933
$begingroup$
it seems good to me.
$endgroup$
– mookid
Mar 25 '14 at 2:15
add a comment |
$begingroup$
it seems good to me.
$endgroup$
– mookid
Mar 25 '14 at 2:15
$begingroup$
it seems good to me.
$endgroup$
– mookid
Mar 25 '14 at 2:15
$begingroup$
it seems good to me.
$endgroup$
– mookid
Mar 25 '14 at 2:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Don't work too hard: $|x| - |y| geq -|x - y|$ is true because $|x| + |y - x| geq |y|$,
and $|x| - |y| leq |x - y|$ because $|y| + |x - y| geq |x|$ . These are true because of triangle inequality, hence the answer.
edited Mar 25 '14 at 2:16
Display Name
803933
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
$endgroup$
3
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
add a comment |
$begingroup$
In $mathbb{R}$ is, for $mathbb{R}^n$, how about
begin{equation}
|x| le |x-y| +|y|.
end{equation}
Analogously,
begin{equation}
|y|-|x| le |x-y|.
end{equation}
or
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$
answered May 2 '14 at 21:47
user29999user29999
3,4601529
$endgroup$
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Don't work too hard: $|x| - |y| geq -|x - y|$ is true because $|x| + |y - x| geq |y|$,
and $|x| - |y| leq |x - y|$ because $|y| + |x - y| geq |x|$ . These are true because of triangle inequality, hence the answer.
edited Mar 25 '14 at 2:16
Display Name
803933
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
$endgroup$
3
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
add a comment |
$begingroup$
Don't work too hard: $|x| - |y| geq -|x - y|$ is true because $|x| + |y - x| geq |y|$,
and $|x| - |y| leq |x - y|$ because $|y| + |x - y| geq |x|$ . These are true because of triangle inequality, hence the answer.
edited Mar 25 '14 at 2:16
Display Name
803933
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
$endgroup$
3
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
add a comment |
$begingroup$
Don't work too hard: $|x| - |y| geq -|x - y|$ is true because $|x| + |y - x| geq |y|$,
and $|x| - |y| leq |x - y|$ because $|y| + |x - y| geq |x|$ . These are true because of triangle inequality, hence the answer.
edited Mar 25 '14 at 2:16
Display Name
803933
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
$endgroup$
Don't work too hard: $|x| - |y| geq -|x - y|$ is true because $|x| + |y - x| geq |y|$,
and $|x| - |y| leq |x - y|$ because $|y| + |x - y| geq |x|$ . These are true because of triangle inequality, hence the answer.
edited Mar 25 '14 at 2:16
Display Name
803933
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
edited Mar 25 '14 at 2:16
Display Name
803933
edited Mar 25 '14 at 2:16
Display Name
803933
edited Mar 25 '14 at 2:16
Display Name
803933
803933
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
answered Mar 25 '14 at 2:14
DeepSeaDeepSea
71.4k54488
71.4k54488
3
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
add a comment |
3
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
3
3
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
$begingroup$
Please use $LaTeX$.
$endgroup$
– Pedro Tamaroff♦
Mar 25 '14 at 2:16
add a comment |
$begingroup$
In $mathbb{R}$ is, for $mathbb{R}^n$, how about
begin{equation}
|x| le |x-y| +|y|.
end{equation}
Analogously,
begin{equation}
|y|-|x| le |x-y|.
end{equation}
or
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$
answered May 2 '14 at 21:47
user29999user29999
3,4601529
$endgroup$
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
add a comment |
$begingroup$
In $mathbb{R}$ is, for $mathbb{R}^n$, how about
begin{equation}
|x| le |x-y| +|y|.
end{equation}
Analogously,
begin{equation}
|y|-|x| le |x-y|.
end{equation}
or
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$
answered May 2 '14 at 21:47
user29999user29999
3,4601529
$endgroup$
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
add a comment |
$begingroup$
In $mathbb{R}$ is, for $mathbb{R}^n$, how about
begin{equation}
|x| le |x-y| +|y|.
end{equation}
Analogously,
begin{equation}
|y|-|x| le |x-y|.
end{equation}
or
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$
answered May 2 '14 at 21:47
user29999user29999
3,4601529
$endgroup$
In $mathbb{R}$ is, for $mathbb{R}^n$, how about
begin{equation}
|x| le |x-y| +|y|.
end{equation}
Analogously,
begin{equation}
|y|-|x| le |x-y|.
end{equation}
or
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$
answered May 2 '14 at 21:47
user29999user29999
3,4601529
answered May 2 '14 at 21:47
user29999user29999
3,4601529
answered May 2 '14 at 21:47
user29999user29999
3,4601529
answered May 2 '14 at 21:47
user29999user29999
3,4601529
3,4601529
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
add a comment |
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Note that this question is older. I doubt this will be of any use to the OP, although it may be of use to other users?
$endgroup$
– Chris K
May 2 '14 at 21:51
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Isn't this exactly my proof in the question? Or are you saying this is the version for $mathbb{R}^n$?
$endgroup$
– Display Name
May 3 '14 at 0:36
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
$begingroup$
Yes, your proof is right and I'm saying that one usual notation for the scalar product in $ mathbb{R}^n$ is $langle x, y rangle $ you may also use $ x cdot y$. If you do not do this people could think that you are proving the real case.
$endgroup$
– user29999
May 3 '14 at 22:43
add a comment |
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$begingroup$
it seems good to me.
$endgroup$
– mookid
Mar 25 '14 at 2:15