Proof of the 'second' triangle inequality












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    it seems good to me.
    $endgroup$
    – mookid
    Mar 25 '14 at 2:15
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    it seems good to me.
    $endgroup$
    – mookid
    Mar 25 '14 at 2:15














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Please use $LaTeX$.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 25 '14 at 2:16



















0












$begingroup$

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
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Please use $LaTeX$.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 25 '14 at 2:16
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Please use $LaTeX$.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 25 '14 at 2:16














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 25 '14 at 2:16









Display Name

803933




803933










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














  • 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











0












$begingroup$

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
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$






share|cite|improve this answer









$endgroup$



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
$$|x-y|^2 = langle x-y,x-yrangle = |x|^2 - 2langle x, y rangle + |y|^2 geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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