Simple two variable am-gm inequality [duplicate]
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$
This inequality seems very easy, I'm feeling dumb for not having solved it yet
inequality a.m.-g.m.-inequality
asked Dec 16 at 10:56
Spasoje Durovic
19410
marked as duplicate by Martin R, Davide Giraudo, Did, kjetil b halvorsen, RRL Dec 16 at 14:13
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.
add a comment |
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$
This inequality seems very easy, I'm feeling dumb for not having solved it yet
inequality a.m.-g.m.-inequality
asked Dec 16 at 10:56
Spasoje Durovic
19410
marked as duplicate by Martin R, Davide Giraudo, Did, kjetil b halvorsen, RRL Dec 16 at 14:13
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.
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
Dec 16 at 10:58
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
Dec 16 at 11:07
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
Dec 16 at 11:11
add a comment |
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$
This inequality seems very easy, I'm feeling dumb for not having solved it yet
inequality a.m.-g.m.-inequality
asked Dec 16 at 10:56
Spasoje Durovic
19410
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Given $x,y in Bbb{R}$, show that:$$x^2+y^2+1ge xy+y+x $$
I tried using the fact that $x^2+y^2 ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1ge x+y$
This inequality seems very easy, I'm feeling dumb for not having solved it yet
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
inequality a.m.-g.m.-inequality
inequality a.m.-g.m.-inequality
asked Dec 16 at 10:56
Spasoje Durovic
19410
asked Dec 16 at 10:56
Spasoje Durovic
19410
asked Dec 16 at 10:56
Spasoje Durovic
19410
asked Dec 16 at 10:56
Spasoje Durovic
19410
asked Dec 16 at 10:56
Spasoje Durovic
19410
19410
marked as duplicate by Martin R, Davide Giraudo, Did, kjetil b halvorsen, RRL Dec 16 at 14:13
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 Martin R, Davide Giraudo, Did, kjetil b halvorsen, RRL Dec 16 at 14:13
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.
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
Dec 16 at 10:58
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
Dec 16 at 11:07
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
Dec 16 at 11:11
add a comment |
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
Dec 16 at 10:58
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
Dec 16 at 11:07
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
Dec 16 at 11:11
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
Dec 16 at 10:58
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
Dec 16 at 10:58
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
Dec 16 at 11:07
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
Dec 16 at 11:07
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
Dec 16 at 11:11
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
Dec 16 at 11:11
add a comment |
3 Answers
3
active
oldest
votes
Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
$$x^2+1geq 2x$$
$$y^2+1geq 2y$$
Now add all these...
answered Dec 16 at 11:00
greedoid
37.7k114794
add a comment |
Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
add a comment |
We need to prove that
$$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
$$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
$$x^2+1geq 2x$$
$$y^2+1geq 2y$$
Now add all these...
answered Dec 16 at 11:00
greedoid
37.7k114794
add a comment |
Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
$$x^2+1geq 2x$$
$$y^2+1geq 2y$$
Now add all these...
answered Dec 16 at 11:00
greedoid
37.7k114794
add a comment |
Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
$$x^2+1geq 2x$$
$$y^2+1geq 2y$$
Now add all these...
answered Dec 16 at 11:00
greedoid
37.7k114794
Since $$a^2-2ab+b^2 = (a-b)^2geq 0implies a^2+b^2geq 2ab$$ we have $$ x^2+y^2geq 2xy$$
$$x^2+1geq 2x$$
$$y^2+1geq 2y$$
Now add all these...
answered Dec 16 at 11:00
greedoid
37.7k114794
answered Dec 16 at 11:00
greedoid
37.7k114794
answered Dec 16 at 11:00
greedoid
37.7k114794
answered Dec 16 at 11:00
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
add a comment |
Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
add a comment |
Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
Hint: Use that $$x^2+y^2+z^2geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$
this is $$(x-y)^2+(y-z)^2+(z-x)^2geq 0$$
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
answered Dec 16 at 10:58
Dr. Sonnhard Graubner
72.9k42865
72.9k42865
add a comment |
add a comment |
We need to prove that
$$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
$$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
add a comment |
We need to prove that
$$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
$$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
add a comment |
We need to prove that
$$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
$$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
We need to prove that
$$y^2-(x+1)y+x^2-x+1geq0,$$ for which it's enough to prove that
$$(x+1)^2-4(x^2-x+1)leq0$$ or $$(x-1)^2geq0.$$
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
answered Dec 16 at 11:07
Michael Rozenberg
95.5k1588184
95.5k1588184
add a comment |
add a comment |
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
Dec 16 at 10:58
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
Dec 16 at 11:07
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
Dec 16 at 11:11