$||x|-|y||leq|x-y|Rightarrow |x-y|geq|x|-|y|$ and $|x+y|geq|x|-|y|$.
$begingroup$
I think I can prove the inequality but in order to do so I Need to understand whether if
$|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$
My proof would be then
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$
And then one can choose for $y$ its negative value and would get
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$
If my idea is Right please help me to prove $(*)$
Otherwise I would like a hint so I can find it out myself
inequality absolute-value
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
asked Jan 8 at 8:36
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
I think I can prove the inequality but in order to do so I Need to understand whether if
$|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$
My proof would be then
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$
And then one can choose for $y$ its negative value and would get
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$
If my idea is Right please help me to prove $(*)$
Otherwise I would like a hint so I can find it out myself
inequality absolute-value
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
asked Jan 8 at 8:36
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
I think I can prove the inequality but in order to do so I Need to understand whether if
$|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$
My proof would be then
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$
And then one can choose for $y$ its negative value and would get
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$
If my idea is Right please help me to prove $(*)$
Otherwise I would like a hint so I can find it out myself
inequality absolute-value
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
asked Jan 8 at 8:36
RM777RM777
38312
$endgroup$
I think I can prove the inequality but in order to do so I Need to understand whether if
$|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$
My proof would be then
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$
And then one can choose for $y$ its negative value and would get
$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$
If my idea is Right please help me to prove $(*)$
Otherwise I would like a hint so I can find it out myself
inequality absolute-value
inequality absolute-value
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
asked Jan 8 at 8:36
RM777RM777
38312
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
asked Jan 8 at 8:36
RM777RM777
38312
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
edited Jan 8 at 10:25
Michael Rozenberg
111k1897201
111k1897201
asked Jan 8 at 8:36
RM777RM777
38312
asked Jan 8 at 8:36
RM777RM777
38312
asked Jan 8 at 8:36
RM777RM777
38312
38312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.
Hence $|a| > |b| ge b$, that is we have $|a| > b$.
Similarly for $-b$.
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Because by the triangle inequality
$$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
$$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
$$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
$$|x+y|geq|x|-|y|.$$
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.
Hence $|a| > |b| ge b$, that is we have $|a| > b$.
Similarly for $-b$.
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.
Hence $|a| > |b| ge b$, that is we have $|a| > b$.
Similarly for $-b$.
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.
Hence $|a| > |b| ge b$, that is we have $|a| > b$.
Similarly for $-b$.
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.
Hence $|a| > |b| ge b$, that is we have $|a| > b$.
Similarly for $-b$.
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
answered Jan 8 at 8:38
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
add a comment |
add a comment |
$begingroup$
Because by the triangle inequality
$$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
$$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
$$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
$$|x+y|geq|x|-|y|.$$
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
Because by the triangle inequality
$$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
$$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
$$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
$$|x+y|geq|x|-|y|.$$
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
Because by the triangle inequality
$$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
$$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
$$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
$$|x+y|geq|x|-|y|.$$
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
Because by the triangle inequality
$$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
$$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
$$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
$$|x+y|geq|x|-|y|.$$
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
edited Jan 8 at 9:35
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 8 at 8:40
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
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