Is this inequality true? $u^2+v^2+s^2+t^2geq (u+v)(s+t)$
$begingroup$
Is this inequality true in $mathbb{R}$? $$u^2+v^2+s^2+t^2geq (u+v)(s+t)$$
I don't know if this is a well-known result. If you have a counterexample or a relevant reference I would appreciate it.
inequality reference-request a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
$endgroup$
add a comment |
$begingroup$
Is this inequality true in $mathbb{R}$? $$u^2+v^2+s^2+t^2geq (u+v)(s+t)$$
I don't know if this is a well-known result. If you have a counterexample or a relevant reference I would appreciate it.
inequality reference-request a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
$endgroup$
add a comment |
$begingroup$
Is this inequality true in $mathbb{R}$? $$u^2+v^2+s^2+t^2geq (u+v)(s+t)$$
I don't know if this is a well-known result. If you have a counterexample or a relevant reference I would appreciate it.
inequality reference-request a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
$endgroup$
Is this inequality true in $mathbb{R}$? $$u^2+v^2+s^2+t^2geq (u+v)(s+t)$$
I don't know if this is a well-known result. If you have a counterexample or a relevant reference I would appreciate it.
inequality reference-request a.m.-g.m.-inequality cauchy-schwarz-inequality
inequality reference-request a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
edited Jan 21 at 12:52
Michael Rozenberg
104k1892197
104k1892197
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
asked Jan 20 at 18:59
ChiloteChilote
1,98811035
1,98811035
add a comment |
add a comment |
3 Answers
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$begingroup$
begin{eqnarray*}
(u-s)^2+ (u-t)^2+(v-s)^2+(v-t)^2geq 0.
end{eqnarray*}
Now divide by $2$ and rearrange.
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
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$begingroup$
Yes, it's true.
By C-S and AM-GM we obtain:
$$u^2+v^2+s^2+t^2=frac{1}{2}left((1^2+1^2)(u^2+v^2)+(1^2+1^2)(s^2+t^2)right)geq$$
$$geqfrac{1}{2}left((u+v)^2+(s+t)^2right)geqsqrt{(u+v)^2(s+t)^2}=|(u+v)(s+t)|geq(u+v)(s+t).$$
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
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add a comment |
$begingroup$
It is $$(u-v)^2+(v-s)^2+(u-t)^2+(v-t)^2geq 0$$
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
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2
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
add a comment |
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3 Answers
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3 Answers
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$begingroup$
begin{eqnarray*}
(u-s)^2+ (u-t)^2+(v-s)^2+(v-t)^2geq 0.
end{eqnarray*}
Now divide by $2$ and rearrange.
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
add a comment |
$begingroup$
begin{eqnarray*}
(u-s)^2+ (u-t)^2+(v-s)^2+(v-t)^2geq 0.
end{eqnarray*}
Now divide by $2$ and rearrange.
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
add a comment |
$begingroup$
begin{eqnarray*}
(u-s)^2+ (u-t)^2+(v-s)^2+(v-t)^2geq 0.
end{eqnarray*}
Now divide by $2$ and rearrange.
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
begin{eqnarray*}
(u-s)^2+ (u-t)^2+(v-s)^2+(v-t)^2geq 0.
end{eqnarray*}
Now divide by $2$ and rearrange.
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
answered Jan 20 at 19:03
Donald SplutterwitDonald Splutterwit
22.8k21445
22.8k21445
add a comment |
add a comment |
$begingroup$
Yes, it's true.
By C-S and AM-GM we obtain:
$$u^2+v^2+s^2+t^2=frac{1}{2}left((1^2+1^2)(u^2+v^2)+(1^2+1^2)(s^2+t^2)right)geq$$
$$geqfrac{1}{2}left((u+v)^2+(s+t)^2right)geqsqrt{(u+v)^2(s+t)^2}=|(u+v)(s+t)|geq(u+v)(s+t).$$
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
Yes, it's true.
By C-S and AM-GM we obtain:
$$u^2+v^2+s^2+t^2=frac{1}{2}left((1^2+1^2)(u^2+v^2)+(1^2+1^2)(s^2+t^2)right)geq$$
$$geqfrac{1}{2}left((u+v)^2+(s+t)^2right)geqsqrt{(u+v)^2(s+t)^2}=|(u+v)(s+t)|geq(u+v)(s+t).$$
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
Yes, it's true.
By C-S and AM-GM we obtain:
$$u^2+v^2+s^2+t^2=frac{1}{2}left((1^2+1^2)(u^2+v^2)+(1^2+1^2)(s^2+t^2)right)geq$$
$$geqfrac{1}{2}left((u+v)^2+(s+t)^2right)geqsqrt{(u+v)^2(s+t)^2}=|(u+v)(s+t)|geq(u+v)(s+t).$$
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
Yes, it's true.
By C-S and AM-GM we obtain:
$$u^2+v^2+s^2+t^2=frac{1}{2}left((1^2+1^2)(u^2+v^2)+(1^2+1^2)(s^2+t^2)right)geq$$
$$geqfrac{1}{2}left((u+v)^2+(s+t)^2right)geqsqrt{(u+v)^2(s+t)^2}=|(u+v)(s+t)|geq(u+v)(s+t).$$
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 20 at 19:03
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
add a comment |
add a comment |
$begingroup$
It is $$(u-v)^2+(v-s)^2+(u-t)^2+(v-t)^2geq 0$$
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
$endgroup$
2
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
add a comment |
$begingroup$
It is $$(u-v)^2+(v-s)^2+(u-t)^2+(v-t)^2geq 0$$
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
$endgroup$
2
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
add a comment |
$begingroup$
It is $$(u-v)^2+(v-s)^2+(u-t)^2+(v-t)^2geq 0$$
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
$endgroup$
It is $$(u-v)^2+(v-s)^2+(u-t)^2+(v-t)^2geq 0$$
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
answered Jan 20 at 19:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
75.9k42866
2
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
add a comment |
2
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
2
2
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
$begingroup$
Are you sure, Sonnhard?
$endgroup$
– Michael Rozenberg
Jan 20 at 21:57
add a comment |
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