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Show that for all $u,v in mathbb{R}, u>v$ if and only if $u^3 > v^3$. [duplicate]
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This question already has an answer here:
$ x^3<y^3 iff x<y $ proof issue.
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Show that for all $u,v in mathbb{R}, u>v$ if and only if $u^3 > v^3$.
Hi, just wondering how everyone thinks best to prove this? I was thinking of using cases but not sure!
Thank you!
algebra-precalculus
edited Jan 28 at 18:13
kelalaka
3351314
asked Jan 28 at 17:51
James odareJames odare
738
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marked as duplicate by Andrés E. Caicedo, Martin R, Lord Shark the Unknown, Maria Mazur algebra-precalculus
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Jan 28 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
$ x^3<y^3 iff x<y $ proof issue.
2 answers
Show that for all $u,v in mathbb{R}, u>v$ if and only if $u^3 > v^3$.
Hi, just wondering how everyone thinks best to prove this? I was thinking of using cases but not sure!
Thank you!
algebra-precalculus
edited Jan 28 at 18:13
kelalaka
3351314
asked Jan 28 at 17:51
James odareJames odare
738
$endgroup$
marked as duplicate by Andrés E. Caicedo, Martin R, Lord Shark the Unknown, Maria Mazur algebra-precalculus
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Jan 28 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Possible duplicate of $ x^3<y^3 iff x<y $ proof issue.
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– Martin R
Jan 28 at 18:23
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$begingroup$
This question already has an answer here:
$ x^3<y^3 iff x<y $ proof issue.
2 answers
Show that for all $u,v in mathbb{R}, u>v$ if and only if $u^3 > v^3$.
Hi, just wondering how everyone thinks best to prove this? I was thinking of using cases but not sure!
Thank you!
algebra-precalculus
edited Jan 28 at 18:13
kelalaka
3351314
asked Jan 28 at 17:51
James odareJames odare
738
$endgroup$
This question already has an answer here:
$ x^3<y^3 iff x<y $ proof issue.
2 answers
Show that for all $u,v in mathbb{R}, u>v$ if and only if $u^3 > v^3$.
Hi, just wondering how everyone thinks best to prove this? I was thinking of using cases but not sure!
Thank you!
This question already has an answer here:
$ x^3<y^3 iff x<y $ proof issue.
2 answers
algebra-precalculus
algebra-precalculus
edited Jan 28 at 18:13
kelalaka
3351314
asked Jan 28 at 17:51
James odareJames odare
738
edited Jan 28 at 18:13
kelalaka
3351314
asked Jan 28 at 17:51
James odareJames odare
738
edited Jan 28 at 18:13
kelalaka
3351314
edited Jan 28 at 18:13
kelalaka
3351314
edited Jan 28 at 18:13
kelalaka
3351314
3351314
asked Jan 28 at 17:51
James odareJames odare
738
asked Jan 28 at 17:51
James odareJames odare
738
asked Jan 28 at 17:51
James odareJames odare
738
738
marked as duplicate by Andrés E. Caicedo, Martin R, Lord Shark the Unknown, Maria Mazur algebra-precalculus
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Jan 28 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Andrés E. Caicedo, Martin R, Lord Shark the Unknown, Maria Mazur algebra-precalculus
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Jan 28 at 18:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
$begingroup$
Possible duplicate of $ x^3<y^3 iff x<y $ proof issue.
$endgroup$
– Martin R
Jan 28 at 18:23
add a comment |
4
$begingroup$
Possible duplicate of $ x^3<y^3 iff x<y $ proof issue.
$endgroup$
– Martin R
Jan 28 at 18:23
4
4
$begingroup$
Possible duplicate of $ x^3<y^3 iff x<y $ proof issue.
$endgroup$
– Martin R
Jan 28 at 18:23
$begingroup$
Possible duplicate of $ x^3<y^3 iff x<y $ proof issue.
$endgroup$
– Martin R
Jan 28 at 18:23
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1 Answer
1
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2
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Hint: Use that $$a^3-b^3=(a-b) left(a^2+a b+b^2right)$$
It is $$a^2+b^2geq 2|ab|$$ by AM.GM, so $$a^2+b^2+abgeq 2|ab|+abgeq 0$$ since $$2|ab|geq -ab$$
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Hint: Use that $$a^3-b^3=(a-b) left(a^2+a b+b^2right)$$
It is $$a^2+b^2geq 2|ab|$$ by AM.GM, so $$a^2+b^2+abgeq 2|ab|+abgeq 0$$ since $$2|ab|geq -ab$$
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
|
show 3 more comments
2
$begingroup$
Hint: Use that $$a^3-b^3=(a-b) left(a^2+a b+b^2right)$$
It is $$a^2+b^2geq 2|ab|$$ by AM.GM, so $$a^2+b^2+abgeq 2|ab|+abgeq 0$$ since $$2|ab|geq -ab$$
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
|
show 3 more comments
2
2
2
$begingroup$
Hint: Use that $$a^3-b^3=(a-b) left(a^2+a b+b^2right)$$
It is $$a^2+b^2geq 2|ab|$$ by AM.GM, so $$a^2+b^2+abgeq 2|ab|+abgeq 0$$ since $$2|ab|geq -ab$$
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
Hint: Use that $$a^3-b^3=(a-b) left(a^2+a b+b^2right)$$
It is $$a^2+b^2geq 2|ab|$$ by AM.GM, so $$a^2+b^2+abgeq 2|ab|+abgeq 0$$ since $$2|ab|geq -ab$$
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
edited Jan 28 at 18:04
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered Jan 28 at 17:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
78.2k42867
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
|
show 3 more comments
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
Thank you, would it be wise to factories the quadratic further or am I missing something?
$endgroup$
– James odare
Jan 28 at 17:56
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
We have $$a^2+ab+b^2geq 0$$ for all real $a,b$
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 17:58
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
I think you don't need cases, you need the factorization as i have written
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:04
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
How did you get the second line? I.E where is the 2 from
$endgroup$
– James odare
Jan 28 at 18:07
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
$begingroup$
I have used the AM-GM inequality $$a+bgeq 2ab$$ for $$a,bgeq 0$$ and then i added $ab$ on both sides.
$endgroup$
– Dr. Sonnhard Graubner
Jan 28 at 18:10
|
show 3 more comments
4
$begingroup$
Possible duplicate of $ x^3<y^3 iff x<y $ proof issue.
$endgroup$
– Martin R
Jan 28 at 18:23