Simple two variable am-gm inequality
up vote
2
down vote
favorite
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 1 hour ago
Spasoje Durovic
1809
add a comment |
up vote
2
down vote
favorite
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 1 hour ago
Spasoje Durovic
1809
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago
Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 1 hour ago
Spasoje Durovic
1809
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
inequality a.m.-g.m.-inequality
asked 1 hour ago
Spasoje Durovic
1809
asked 1 hour ago
Spasoje Durovic
1809
asked 1 hour ago
Spasoje Durovic
1809
asked 1 hour ago
Spasoje Durovic
1809
asked 1 hour ago
Spasoje Durovic
1809
1809
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago
Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago
add a comment |
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago
Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago
Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago
Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago
Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
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 1 hour ago
greedoid
36.6k114592
add a comment |
up vote
5
down vote
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 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
add a comment |
up vote
2
down vote
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 1 hour ago
Michael Rozenberg
95k1588183
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
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 1 hour ago
greedoid
36.6k114592
add a comment |
up vote
7
down vote
accepted
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 1 hour ago
greedoid
36.6k114592
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
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 1 hour ago
greedoid
36.6k114592
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 1 hour ago
greedoid
36.6k114592
answered 1 hour ago
greedoid
36.6k114592
answered 1 hour ago
greedoid
36.6k114592
answered 1 hour ago
greedoid
36.6k114592
36.6k114592
add a comment |
add a comment |
up vote
5
down vote
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 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
add a comment |
up vote
5
down vote
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 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
add a comment |
up vote
5
down vote
up vote
5
down vote
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 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
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 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
answered 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
answered 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
answered 1 hour ago
Dr. Sonnhard Graubner
72.4k32865
72.4k32865
add a comment |
add a comment |
up vote
2
down vote
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 1 hour ago
Michael Rozenberg
95k1588183
add a comment |
up vote
2
down vote
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 1 hour ago
Michael Rozenberg
95k1588183
add a comment |
up vote
2
down vote
up vote
2
down vote
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 1 hour ago
Michael Rozenberg
95k1588183
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 1 hour ago
Michael Rozenberg
95k1588183
answered 1 hour ago
Michael Rozenberg
95k1588183
answered 1 hour ago
Michael Rozenberg
95k1588183
answered 1 hour ago
Michael Rozenberg
95k1588183
95k1588183
add a comment |
add a comment |
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Fix $y$, differentiate w.r.t. $x$, etc
– mathworker21
1 hour ago
Possible duplicate of How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$? – Also: math.stackexchange.com/q/772220/42969
– Martin R
1 hour ago
@MartinR I just didn't see them being equivalent, I well new that one inequality
– Spasoje Durovic
1 hour ago
@Spasoje Durovic Substitute $c=1$.
– Michael Rozenberg
57 mins ago