Mixing
Reverse Triangle Inequality Proof
$begingroup$
I've seen the full proof of the Triangle Inequality
begin{equation*}
|x+y|le|x|+|y|.
end{equation*}
However, I haven't seen the proof of the reverse triangle inequality:
begin{equation*}
||x|-|y||le|x-y|.
end{equation*}
Would you please prove this using only the Triangle Inequality above?
Thank you very much.
real-analysis inequality absolute-value
edited May 26 '15 at 19:12
George Simpson
4,57432449
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
$endgroup$
add a comment |
$begingroup$
I've seen the full proof of the Triangle Inequality
begin{equation*}
|x+y|le|x|+|y|.
end{equation*}
However, I haven't seen the proof of the reverse triangle inequality:
begin{equation*}
||x|-|y||le|x-y|.
end{equation*}
Would you please prove this using only the Triangle Inequality above?
Thank you very much.
real-analysis inequality absolute-value
edited May 26 '15 at 19:12
George Simpson
4,57432449
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
$endgroup$
3
$begingroup$
proofwiki.org/wiki/Reverse_Triangle_Inequality
$endgroup$
– dls
Apr 2 '12 at 20:15
1
$begingroup$
I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.
$endgroup$
– Anonymous
Apr 2 '12 at 20:16
12
$begingroup$
Just replace $d(x,y)$ with $|x-y|$.
$endgroup$
– dls
Apr 2 '12 at 20:21
$begingroup$
this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?
$endgroup$
– Charlie Parker
Feb 14 '18 at 16:24
$begingroup$
If you think about $x$ and $y$ as points in $mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).
$endgroup$
– user428487
Mar 27 '18 at 0:50
add a comment |
$begingroup$
I've seen the full proof of the Triangle Inequality
begin{equation*}
|x+y|le|x|+|y|.
end{equation*}
However, I haven't seen the proof of the reverse triangle inequality:
begin{equation*}
||x|-|y||le|x-y|.
end{equation*}
Would you please prove this using only the Triangle Inequality above?
Thank you very much.
real-analysis inequality absolute-value
edited May 26 '15 at 19:12
George Simpson
4,57432449
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
$endgroup$
I've seen the full proof of the Triangle Inequality
begin{equation*}
|x+y|le|x|+|y|.
end{equation*}
However, I haven't seen the proof of the reverse triangle inequality:
begin{equation*}
||x|-|y||le|x-y|.
end{equation*}
Would you please prove this using only the Triangle Inequality above?
Thank you very much.
real-analysis inequality absolute-value
real-analysis inequality absolute-value
edited May 26 '15 at 19:12
George Simpson
4,57432449
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
edited May 26 '15 at 19:12
George Simpson
4,57432449
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
edited May 26 '15 at 19:12
George Simpson
4,57432449
edited May 26 '15 at 19:12
George Simpson
4,57432449
edited May 26 '15 at 19:12
George Simpson
4,57432449
4,57432449
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
asked Apr 2 '12 at 20:10
AnonymousAnonymous
9841818
9841818
3
$begingroup$
proofwiki.org/wiki/Reverse_Triangle_Inequality
$endgroup$
– dls
Apr 2 '12 at 20:15
1
$begingroup$
I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.
$endgroup$
– Anonymous
Apr 2 '12 at 20:16
12
$begingroup$
Just replace $d(x,y)$ with $|x-y|$.
$endgroup$
– dls
Apr 2 '12 at 20:21
$begingroup$
this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?
$endgroup$
– Charlie Parker
Feb 14 '18 at 16:24
$begingroup$
If you think about $x$ and $y$ as points in $mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).
$endgroup$
– user428487
Mar 27 '18 at 0:50
add a comment |
3
$begingroup$
proofwiki.org/wiki/Reverse_Triangle_Inequality
$endgroup$
– dls
Apr 2 '12 at 20:15
1
$begingroup$
I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.
$endgroup$
– Anonymous
Apr 2 '12 at 20:16
12
$begingroup$
Just replace $d(x,y)$ with $|x-y|$.
$endgroup$
– dls
Apr 2 '12 at 20:21
$begingroup$
this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?
$endgroup$
– Charlie Parker
Feb 14 '18 at 16:24
$begingroup$
If you think about $x$ and $y$ as points in $mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).
$endgroup$
– user428487
Mar 27 '18 at 0:50
3
3
$begingroup$
proofwiki.org/wiki/Reverse_Triangle_Inequality
$endgroup$
– dls
Apr 2 '12 at 20:15
$begingroup$
proofwiki.org/wiki/Reverse_Triangle_Inequality
$endgroup$
– dls
Apr 2 '12 at 20:15
1
1
$begingroup$
I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.
$endgroup$
– Anonymous
Apr 2 '12 at 20:16
$begingroup$
I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.
$endgroup$
– Anonymous
Apr 2 '12 at 20:16
12
12
$begingroup$
Just replace $d(x,y)$ with $|x-y|$.
$endgroup$
– dls
Apr 2 '12 at 20:21
$begingroup$
Just replace $d(x,y)$ with $|x-y|$.
$endgroup$
– dls
Apr 2 '12 at 20:21
$begingroup$
this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?
$endgroup$
– Charlie Parker
Feb 14 '18 at 16:24
$begingroup$
this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?
$endgroup$
– Charlie Parker
Feb 14 '18 at 16:24
$begingroup$
If you think about $x$ and $y$ as points in $mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).
$endgroup$
– user428487
Mar 27 '18 at 0:50
$begingroup$
If you think about $x$ and $y$ as points in $mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).
$endgroup$
– user428487
Mar 27 '18 at 0:50
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$$|x| + |y -x| ge |x + y -x| = |y|$$
$$|y| + |x -y| ge |y + x -y| = |x|$$
Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get:
$$|y -x| ge |y| - |x|$$
$$|x -y| ge |x| -|y|.$$
From absolute value properties we know that $|y-x| = |x-y|$ and if $t ge a$ and $t ge −a$ then $t ge |a|$.
Combining these two facts together we get the reverse triangle inequality:
$$|x-y| ge bigl||x|-|y|bigr|.$$
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
$endgroup$
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
5
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
1
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
2
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
|
show 13 more comments
$begingroup$
WLOG, consider $|x|ge |y|$. Hence:
$$||x|-|y||=||x-y+y|-|y||le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
$endgroup$
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
add a comment |
$begingroup$
Given that we are discussing the reals, $mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,zinmathbb{R}, quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$.
Then apply $|x| = |(x-y)+y|leq |x-y|+|y|$. By so-called "first triangle inequality."
Rewriting $|x|-|y| leq |x-y|$ and $||x|-y|| leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(leq,geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers.
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
$endgroup$
6
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
add a comment |
$begingroup$
Explicitly, we have
begin{equation}
bigl||x|-|y|bigr|
=
left{
begin{array}{ll}
|x-y|=x-y,&xgeq{}ygeq0\
|x-y|=-x+y=-(x-y),&ygeq{}xgeq0\
|-x-y|=x+yleq-x+y=-(x-y),&ygeq-xgeq0\
|-x-y|=-x-yleq-x+y=-(x-y),&-xgeq{}ygeq0\
|-x+y|=-x+y=-(x-y),&-xgeq-ygeq0\
|-x+y|=x-y,&-ygeq-xgeq0\
|-x+y|=-x-yleq{}x-y,&-ygeq{}xgeq0\
|-x+y|=x+yleq{}x-y,&xgeq-ygeq0
end{array}
right}
=|x-y|.nonumber
end{equation}
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$|x| + |y -x| ge |x + y -x| = |y|$$
$$|y| + |x -y| ge |y + x -y| = |x|$$
Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get:
$$|y -x| ge |y| - |x|$$
$$|x -y| ge |x| -|y|.$$
From absolute value properties we know that $|y-x| = |x-y|$ and if $t ge a$ and $t ge −a$ then $t ge |a|$.
Combining these two facts together we get the reverse triangle inequality:
$$|x-y| ge bigl||x|-|y|bigr|.$$
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
$endgroup$
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
5
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
1
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
2
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
|
show 13 more comments
$begingroup$
$$|x| + |y -x| ge |x + y -x| = |y|$$
$$|y| + |x -y| ge |y + x -y| = |x|$$
Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get:
$$|y -x| ge |y| - |x|$$
$$|x -y| ge |x| -|y|.$$
From absolute value properties we know that $|y-x| = |x-y|$ and if $t ge a$ and $t ge −a$ then $t ge |a|$.
Combining these two facts together we get the reverse triangle inequality:
$$|x-y| ge bigl||x|-|y|bigr|.$$
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
$endgroup$
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
5
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
1
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
2
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
|
show 13 more comments
$begingroup$
$$|x| + |y -x| ge |x + y -x| = |y|$$
$$|y| + |x -y| ge |y + x -y| = |x|$$
Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get:
$$|y -x| ge |y| - |x|$$
$$|x -y| ge |x| -|y|.$$
From absolute value properties we know that $|y-x| = |x-y|$ and if $t ge a$ and $t ge −a$ then $t ge |a|$.
Combining these two facts together we get the reverse triangle inequality:
$$|x-y| ge bigl||x|-|y|bigr|.$$
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
$endgroup$
$$|x| + |y -x| ge |x + y -x| = |y|$$
$$|y| + |x -y| ge |y + x -y| = |x|$$
Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get:
$$|y -x| ge |y| - |x|$$
$$|x -y| ge |x| -|y|.$$
From absolute value properties we know that $|y-x| = |x-y|$ and if $t ge a$ and $t ge −a$ then $t ge |a|$.
Combining these two facts together we get the reverse triangle inequality:
$$|x-y| ge bigl||x|-|y|bigr|.$$
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
edited Jun 19 '18 at 4:09
user99914
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
answered Apr 2 '12 at 20:15
AryabhataAryabhata
70.2k6156246
70.2k6156246
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
5
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
1
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
2
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
|
show 13 more comments
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
5
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
1
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
2
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:20
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
$begingroup$
Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!
$endgroup$
– Anonymous
Apr 2 '12 at 20:21
5
5
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
$begingroup$
@Anonymous: I suggest you try finishing it yourself first.
$endgroup$
– Aryabhata
Apr 2 '12 at 20:22
1
1
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
$begingroup$
OK now I get it - you say: $|y-x|ge |y|-|x|$ and $|x-y|ge |x|-|y|$ therefore $|x-y|ge ||x|-|y||$
$endgroup$
– Anonymous
Apr 2 '12 at 20:44
2
2
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
$begingroup$
Indeed this shows that the function $x mapsto |x|$ is Lipschitz continuous with constant 1.
$endgroup$
– copper.hat
Apr 2 '12 at 22:07
|
show 13 more comments
$begingroup$
WLOG, consider $|x|ge |y|$. Hence:
$$||x|-|y||=||x-y+y|-|y||le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
$endgroup$
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
add a comment |
$begingroup$
WLOG, consider $|x|ge |y|$. Hence:
$$||x|-|y||=||x-y+y|-|y||le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
$endgroup$
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
add a comment |
$begingroup$
WLOG, consider $|x|ge |y|$. Hence:
$$||x|-|y||=||x-y+y|-|y||le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
$endgroup$
WLOG, consider $|x|ge |y|$. Hence:
$$||x|-|y||=||x-y+y|-|y||le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
edited Sep 6 '18 at 12:56
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
answered Jun 19 '18 at 14:16
farruhotafarruhota
19.7k2738
19.7k2738
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
add a comment |
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $aleq b$, then $left|a-|y|right| leq left|b-|y|right|$.
$endgroup$
– Kenny LJ
Sep 12 '18 at 2:38
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
$begingroup$
Since $|x|ge |y|$, then $||x|-|y||=|x|-|y|ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.
$endgroup$
– farruhota
Sep 12 '18 at 2:43
add a comment |
$begingroup$
Given that we are discussing the reals, $mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,zinmathbb{R}, quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$.
Then apply $|x| = |(x-y)+y|leq |x-y|+|y|$. By so-called "first triangle inequality."
Rewriting $|x|-|y| leq |x-y|$ and $||x|-y|| leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(leq,geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers.
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
$endgroup$
6
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
add a comment |
$begingroup$
Given that we are discussing the reals, $mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,zinmathbb{R}, quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$.
Then apply $|x| = |(x-y)+y|leq |x-y|+|y|$. By so-called "first triangle inequality."
Rewriting $|x|-|y| leq |x-y|$ and $||x|-y|| leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(leq,geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers.
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
$endgroup$
6
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
add a comment |
$begingroup$
Given that we are discussing the reals, $mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,zinmathbb{R}, quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$.
Then apply $|x| = |(x-y)+y|leq |x-y|+|y|$. By so-called "first triangle inequality."
Rewriting $|x|-|y| leq |x-y|$ and $||x|-y|| leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(leq,geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers.
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
$endgroup$
Given that we are discussing the reals, $mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,zinmathbb{R}, quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$.
Then apply $|x| = |(x-y)+y|leq |x-y|+|y|$. By so-called "first triangle inequality."
Rewriting $|x|-|y| leq |x-y|$ and $||x|-y|| leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(leq,geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers.
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
answered Mar 16 '16 at 7:09
Field of NodesField of Nodes
816
816
6
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
add a comment |
6
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
6
6
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
$begingroup$
You can't immediately conclude that $||x|-|y|| leqslant |x-y|$ from $|x|-|y| leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| leqslant |x-y|$ which implies $|x| -|y| geqslant -|x-y|$. Fill in the details and I'll upvote.
$endgroup$
– RRL
Mar 16 '16 at 8:16
add a comment |
$begingroup$
Explicitly, we have
begin{equation}
bigl||x|-|y|bigr|
=
left{
begin{array}{ll}
|x-y|=x-y,&xgeq{}ygeq0\
|x-y|=-x+y=-(x-y),&ygeq{}xgeq0\
|-x-y|=x+yleq-x+y=-(x-y),&ygeq-xgeq0\
|-x-y|=-x-yleq-x+y=-(x-y),&-xgeq{}ygeq0\
|-x+y|=-x+y=-(x-y),&-xgeq-ygeq0\
|-x+y|=x-y,&-ygeq-xgeq0\
|-x+y|=-x-yleq{}x-y,&-ygeq{}xgeq0\
|-x+y|=x+yleq{}x-y,&xgeq-ygeq0
end{array}
right}
=|x-y|.nonumber
end{equation}
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
$endgroup$
add a comment |
$begingroup$
Explicitly, we have
begin{equation}
bigl||x|-|y|bigr|
=
left{
begin{array}{ll}
|x-y|=x-y,&xgeq{}ygeq0\
|x-y|=-x+y=-(x-y),&ygeq{}xgeq0\
|-x-y|=x+yleq-x+y=-(x-y),&ygeq-xgeq0\
|-x-y|=-x-yleq-x+y=-(x-y),&-xgeq{}ygeq0\
|-x+y|=-x+y=-(x-y),&-xgeq-ygeq0\
|-x+y|=x-y,&-ygeq-xgeq0\
|-x+y|=-x-yleq{}x-y,&-ygeq{}xgeq0\
|-x+y|=x+yleq{}x-y,&xgeq-ygeq0
end{array}
right}
=|x-y|.nonumber
end{equation}
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
$endgroup$
add a comment |
$begingroup$
Explicitly, we have
begin{equation}
bigl||x|-|y|bigr|
=
left{
begin{array}{ll}
|x-y|=x-y,&xgeq{}ygeq0\
|x-y|=-x+y=-(x-y),&ygeq{}xgeq0\
|-x-y|=x+yleq-x+y=-(x-y),&ygeq-xgeq0\
|-x-y|=-x-yleq-x+y=-(x-y),&-xgeq{}ygeq0\
|-x+y|=-x+y=-(x-y),&-xgeq-ygeq0\
|-x+y|=x-y,&-ygeq-xgeq0\
|-x+y|=-x-yleq{}x-y,&-ygeq{}xgeq0\
|-x+y|=x+yleq{}x-y,&xgeq-ygeq0
end{array}
right}
=|x-y|.nonumber
end{equation}
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
$endgroup$
Explicitly, we have
begin{equation}
bigl||x|-|y|bigr|
=
left{
begin{array}{ll}
|x-y|=x-y,&xgeq{}ygeq0\
|x-y|=-x+y=-(x-y),&ygeq{}xgeq0\
|-x-y|=x+yleq-x+y=-(x-y),&ygeq-xgeq0\
|-x-y|=-x-yleq-x+y=-(x-y),&-xgeq{}ygeq0\
|-x+y|=-x+y=-(x-y),&-xgeq-ygeq0\
|-x+y|=x-y,&-ygeq-xgeq0\
|-x+y|=-x-yleq{}x-y,&-ygeq{}xgeq0\
|-x+y|=x+yleq{}x-y,&xgeq-ygeq0
end{array}
right}
=|x-y|.nonumber
end{equation}
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
answered Sep 6 '18 at 9:46
bkarpuzbkarpuz
500210
500210
add a comment |
add a comment |
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$begingroup$
proofwiki.org/wiki/Reverse_Triangle_Inequality
$endgroup$
– dls
Apr 2 '12 at 20:15
1
$begingroup$
I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.
$endgroup$
– Anonymous
Apr 2 '12 at 20:16
12
$begingroup$
Just replace $d(x,y)$ with $|x-y|$.
$endgroup$
– dls
Apr 2 '12 at 20:21
$begingroup$
this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?
$endgroup$
– Charlie Parker
Feb 14 '18 at 16:24
$begingroup$
If you think about $x$ and $y$ as points in $mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).
$endgroup$
– user428487
Mar 27 '18 at 0:50