How does $ frac{x^2 + y^2}{2} geq |xy|$ come from $ frac{x + y}{2} geq sqrt{xy}$?
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I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
edited 2 days ago
YuiTo Cheng
2,1862937
asked Mar 27 at 23:43
hopefullyhopefully
279214
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add a comment |
$begingroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
edited 2 days ago
YuiTo Cheng
2,1862937
asked Mar 27 at 23:43
hopefullyhopefully
279214
$endgroup$
add a comment |
$begingroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
edited 2 days ago
YuiTo Cheng
2,1862937
asked Mar 27 at 23:43
hopefullyhopefully
279214
$endgroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
calculus inequality
edited 2 days ago
YuiTo Cheng
2,1862937
asked Mar 27 at 23:43
hopefullyhopefully
279214
edited 2 days ago
YuiTo Cheng
2,1862937
asked Mar 27 at 23:43
hopefullyhopefully
279214
edited 2 days ago
YuiTo Cheng
2,1862937
edited 2 days ago
YuiTo Cheng
2,1862937
edited 2 days ago
YuiTo Cheng
2,1862937
2,1862937
asked Mar 27 at 23:43
hopefullyhopefully
279214
asked Mar 27 at 23:43
hopefullyhopefully
279214
asked Mar 27 at 23:43
hopefullyhopefully
279214
279214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.2k32143
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add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.2k32143
$endgroup$
add a comment |
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.2k32143
$endgroup$
add a comment |
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.2k32143
$endgroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.2k32143
answered Mar 27 at 23:46
jgonjgon
16.2k32143
answered Mar 27 at 23:46
jgonjgon
16.2k32143
answered Mar 27 at 23:46
jgonjgon
16.2k32143
16.2k32143
add a comment |
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
$endgroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
answered Mar 28 at 0:07
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
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