Proof by Deduction $sqrtxy ≤ fracx+y2$
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I want to ask a question about proof of deduction.
I sat my Pure Mathematics Exam more than $3$ years ago but decided to return to the subject for a refresher.
Proofs were not a requirement for my course but as my younger siblings are studying it, I decided to give it a go.
This was the question:
Prove that for all positive values of x and y
$$sqrtxy ≤ fracx+y2$$
Now, I did some research on proofs of deduction and it involved a start point.
My instinct was to work backwards from this inequality to something more meaningful towards this "start point" and work forwards.
This is my working thus far:
$$xy ≤ frac(x+y)^24$$
$$4xy ≤ (x+y)^2$$
$$4xy ≤ x^2 + 2xy + y^2$$
Unfortunately, I can't seem to see where I can go further to start this proof.
Is this the correct approach? If so, is there a further step that I cannot see?
proof-writing
asked Aug 26 at 19:40
Bob Smith
429312
 |Â
show 1 more comment
up vote
1
down vote
favorite
I want to ask a question about proof of deduction.
I sat my Pure Mathematics Exam more than $3$ years ago but decided to return to the subject for a refresher.
Proofs were not a requirement for my course but as my younger siblings are studying it, I decided to give it a go.
This was the question:
Prove that for all positive values of x and y
$$sqrtxy ≤ fracx+y2$$
Now, I did some research on proofs of deduction and it involved a start point.
My instinct was to work backwards from this inequality to something more meaningful towards this "start point" and work forwards.
This is my working thus far:
$$xy ≤ frac(x+y)^24$$
$$4xy ≤ (x+y)^2$$
$$4xy ≤ x^2 + 2xy + y^2$$
Unfortunately, I can't seem to see where I can go further to start this proof.
Is this the correct approach? If so, is there a further step that I cannot see?
proof-writing
asked Aug 26 at 19:40
Bob Smith
429312
2
Write it as $,x-2sqrtxsqrty+y ge 0,$ and recognize a different square on the LHS.
– dxiv
Aug 26 at 19:41
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side.
– Bob Smith
Aug 26 at 19:44
Subtract $4xy$ from both sides of the last inequality.
– saulspatz
Aug 26 at 19:45
2
@BobSmith $sqrtxy ≤ fracx+y2 iff 2sqrtxy le x+y iff 2sqrtxy colorred- 2 sqrtxyle x+ycolorred- 2 sqrtxy,$
– dxiv
Aug 26 at 19:46
1
@dxiv oh. ($sqrtx - sqrty)^2$
– Bob Smith
Aug 26 at 19:50
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to ask a question about proof of deduction.
I sat my Pure Mathematics Exam more than $3$ years ago but decided to return to the subject for a refresher.
Proofs were not a requirement for my course but as my younger siblings are studying it, I decided to give it a go.
This was the question:
Prove that for all positive values of x and y
$$sqrtxy ≤ fracx+y2$$
Now, I did some research on proofs of deduction and it involved a start point.
My instinct was to work backwards from this inequality to something more meaningful towards this "start point" and work forwards.
This is my working thus far:
$$xy ≤ frac(x+y)^24$$
$$4xy ≤ (x+y)^2$$
$$4xy ≤ x^2 + 2xy + y^2$$
Unfortunately, I can't seem to see where I can go further to start this proof.
Is this the correct approach? If so, is there a further step that I cannot see?
proof-writing
asked Aug 26 at 19:40
Bob Smith
429312
I want to ask a question about proof of deduction.
I sat my Pure Mathematics Exam more than $3$ years ago but decided to return to the subject for a refresher.
Proofs were not a requirement for my course but as my younger siblings are studying it, I decided to give it a go.
This was the question:
Prove that for all positive values of x and y
$$sqrtxy ≤ fracx+y2$$
Now, I did some research on proofs of deduction and it involved a start point.
My instinct was to work backwards from this inequality to something more meaningful towards this "start point" and work forwards.
This is my working thus far:
$$xy ≤ frac(x+y)^24$$
$$4xy ≤ (x+y)^2$$
$$4xy ≤ x^2 + 2xy + y^2$$
Unfortunately, I can't seem to see where I can go further to start this proof.
Is this the correct approach? If so, is there a further step that I cannot see?
proof-writing
asked Aug 26 at 19:40
Bob Smith
429312
asked Aug 26 at 19:40
Bob Smith
429312
asked Aug 26 at 19:40
Bob Smith
429312
asked Aug 26 at 19:40
Bob Smith
429312
429312
2
Write it as $,x-2sqrtxsqrty+y ge 0,$ and recognize a different square on the LHS.
– dxiv
Aug 26 at 19:41
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side.
– Bob Smith
Aug 26 at 19:44
Subtract $4xy$ from both sides of the last inequality.
– saulspatz
Aug 26 at 19:45
2
@BobSmith $sqrtxy ≤ fracx+y2 iff 2sqrtxy le x+y iff 2sqrtxy colorred- 2 sqrtxyle x+ycolorred- 2 sqrtxy,$
– dxiv
Aug 26 at 19:46
1
@dxiv oh. ($sqrtx - sqrty)^2$
– Bob Smith
Aug 26 at 19:50
 |Â
show 1 more comment
2
Write it as $,x-2sqrtxsqrty+y ge 0,$ and recognize a different square on the LHS.
– dxiv
Aug 26 at 19:41
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side.
– Bob Smith
Aug 26 at 19:44
Subtract $4xy$ from both sides of the last inequality.
– saulspatz
Aug 26 at 19:45
2
@BobSmith $sqrtxy ≤ fracx+y2 iff 2sqrtxy le x+y iff 2sqrtxy colorred- 2 sqrtxyle x+ycolorred- 2 sqrtxy,$
– dxiv
Aug 26 at 19:46
1
@dxiv oh. ($sqrtx - sqrty)^2$
– Bob Smith
Aug 26 at 19:50
2
2
Write it as $,x-2sqrtxsqrty+y ge 0,$ and recognize a different square on the LHS.
– dxiv
Aug 26 at 19:41
Write it as $,x-2sqrtxsqrty+y ge 0,$ and recognize a different square on the LHS.
– dxiv
Aug 26 at 19:41
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side.
– Bob Smith
Aug 26 at 19:44
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side.
– Bob Smith
Aug 26 at 19:44
Subtract $4xy$ from both sides of the last inequality.
– saulspatz
Aug 26 at 19:45
Subtract $4xy$ from both sides of the last inequality.
– saulspatz
Aug 26 at 19:45
2
2
@BobSmith $sqrtxy ≤ fracx+y2 iff 2sqrtxy le x+y iff 2sqrtxy colorred- 2 sqrtxyle x+ycolorred- 2 sqrtxy,$
– dxiv
Aug 26 at 19:46
@BobSmith $sqrtxy ≤ fracx+y2 iff 2sqrtxy le x+y iff 2sqrtxy colorred- 2 sqrtxyle x+ycolorred- 2 sqrtxy,$
– dxiv
Aug 26 at 19:46
1
1
@dxiv oh. ($sqrtx - sqrty)^2$
– Bob Smith
Aug 26 at 19:50
@dxiv oh. ($sqrtx - sqrty)^2$
– Bob Smith
Aug 26 at 19:50
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
... $rightarrow$ $0le x^2-2xy+y^2=(x-y)^2$
answered Aug 26 at 19:50
Selflearner
263211
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
add a comment |Â
up vote
2
down vote
$sqrtxy ≤ fracx+y2 iff 2sqrtxy ≤ x+y iff x-2sqrtxy+y geq 0$ $ iff (sqrtx-sqrty)^2 geq 0 $.
Which is true
answered Aug 26 at 19:50
valer
3731213
add a comment |Â
up vote
1
down vote
From your last step, you could proceed as follows:
$$
4xy leq x^2 + 2xy + y^2\
0 leq x^2 - 2xy + y^2 = (x-y)^2.
$$
Now you can work backward as you wanted.
answered Aug 26 at 19:51
Brahadeesh
4,11331550
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
... $rightarrow$ $0le x^2-2xy+y^2=(x-y)^2$
answered Aug 26 at 19:50
Selflearner
263211
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
add a comment |Â
up vote
3
down vote
accepted
... $rightarrow$ $0le x^2-2xy+y^2=(x-y)^2$
answered Aug 26 at 19:50
Selflearner
263211
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
... $rightarrow$ $0le x^2-2xy+y^2=(x-y)^2$
answered Aug 26 at 19:50
Selflearner
263211
... $rightarrow$ $0le x^2-2xy+y^2=(x-y)^2$
answered Aug 26 at 19:50
Selflearner
263211
answered Aug 26 at 19:50
Selflearner
263211
answered Aug 26 at 19:50
Selflearner
263211
answered Aug 26 at 19:50
Selflearner
263211
263211
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
add a comment |Â
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
Yes I saw that just now. Thanks!
– Bob Smith
Aug 26 at 19:50
add a comment |Â
up vote
2
down vote
$sqrtxy ≤ fracx+y2 iff 2sqrtxy ≤ x+y iff x-2sqrtxy+y geq 0$ $ iff (sqrtx-sqrty)^2 geq 0 $.
Which is true
answered Aug 26 at 19:50
valer
3731213
add a comment |Â
up vote
2
down vote
$sqrtxy ≤ fracx+y2 iff 2sqrtxy ≤ x+y iff x-2sqrtxy+y geq 0$ $ iff (sqrtx-sqrty)^2 geq 0 $.
Which is true
answered Aug 26 at 19:50
valer
3731213
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$sqrtxy ≤ fracx+y2 iff 2sqrtxy ≤ x+y iff x-2sqrtxy+y geq 0$ $ iff (sqrtx-sqrty)^2 geq 0 $.
Which is true
answered Aug 26 at 19:50
valer
3731213
$sqrtxy ≤ fracx+y2 iff 2sqrtxy ≤ x+y iff x-2sqrtxy+y geq 0$ $ iff (sqrtx-sqrty)^2 geq 0 $.
Which is true
answered Aug 26 at 19:50
valer
3731213
answered Aug 26 at 19:50
valer
3731213
answered Aug 26 at 19:50
valer
3731213
answered Aug 26 at 19:50
valer
3731213
3731213
add a comment |Â
add a comment |Â
up vote
1
down vote
From your last step, you could proceed as follows:
$$
4xy leq x^2 + 2xy + y^2\
0 leq x^2 - 2xy + y^2 = (x-y)^2.
$$
Now you can work backward as you wanted.
answered Aug 26 at 19:51
Brahadeesh
4,11331550
add a comment |Â
up vote
1
down vote
From your last step, you could proceed as follows:
$$
4xy leq x^2 + 2xy + y^2\
0 leq x^2 - 2xy + y^2 = (x-y)^2.
$$
Now you can work backward as you wanted.
answered Aug 26 at 19:51
Brahadeesh
4,11331550
add a comment |Â
up vote
1
down vote
up vote
1
down vote
From your last step, you could proceed as follows:
$$
4xy leq x^2 + 2xy + y^2\
0 leq x^2 - 2xy + y^2 = (x-y)^2.
$$
Now you can work backward as you wanted.
answered Aug 26 at 19:51
Brahadeesh
4,11331550
From your last step, you could proceed as follows:
$$
4xy leq x^2 + 2xy + y^2\
0 leq x^2 - 2xy + y^2 = (x-y)^2.
$$
Now you can work backward as you wanted.
answered Aug 26 at 19:51
Brahadeesh
4,11331550
answered Aug 26 at 19:51
Brahadeesh
4,11331550
answered Aug 26 at 19:51
Brahadeesh
4,11331550
answered Aug 26 at 19:51
Brahadeesh
4,11331550
4,11331550
add a comment |Â
add a comment |Â
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2
Write it as $,x-2sqrtxsqrty+y ge 0,$ and recognize a different square on the LHS.
– dxiv
Aug 26 at 19:41
@dxiv how did you achieve that expression? Is it possible to square root the entire right side and achieve that? I thought you couldn't simplify the root of the right side.
– Bob Smith
Aug 26 at 19:44
Subtract $4xy$ from both sides of the last inequality.
– saulspatz
Aug 26 at 19:45
2
@BobSmith $sqrtxy ≤ fracx+y2 iff 2sqrtxy le x+y iff 2sqrtxy colorred- 2 sqrtxyle x+ycolorred- 2 sqrtxy,$
– dxiv
Aug 26 at 19:46
1
@dxiv oh. ($sqrtx - sqrty)^2$
– Bob Smith
Aug 26 at 19:50