How does $ fracx^2 + y^22 geq |xy|$ come from $ fracx + y2 geq sqrtxy$?How is this form of the Chernoff bound derived from the other?Show that $fracxyz + fracxzy + fracyzx geq x+y+z $ by considering homogeneityHow to get $sqrt k + frac1sqrtk+1$ in the form $fracsqrtk^2 + 1sqrtk+1$?Why does $(x+3)/(x-4) geq 0$ not include 4 in the interval result?Prove $fraca+bsqrtc+fraca+csqrtb+ fracb+csqrta geq 2(sqrta + sqrtb +sqrtc)$Prove $fracsqrta+ sqrtb 2 leq sqrt fraca+b2 $Nesbitt's Inequality $fracab+c+fracbc+a+fracca+bgeqfrac32$Prove when $abc=1$: $ fraca2+bc + fracb2+ca+fracc2+ab geq 1$Where does this trignometric substition come from?Show that $fraca+b2 geq sqrtab geq frac2aba+b$
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How does $ fracx^2 + y^22 geq |xy|$ come from $ fracx + y2 geq sqrtxy$?
How is this form of the Chernoff bound derived from the other?Show that $fracxyz + fracxzy + fracyzx geq x+y+z $ by considering homogeneityHow to get $sqrt k + frac1sqrtk+1$ in the form $fracsqrtk^2 + 1sqrtk+1$?Why does $(x+3)/(x-4) geq 0$ not include 4 in the interval result?Prove $fraca+bsqrtc+fraca+csqrtb+ fracb+csqrta geq 2(sqrta + sqrtb +sqrtc)$Prove $fracsqrta+ sqrtb 2 leq sqrt fraca+b2 $Nesbitt's Inequality $fracab+c+fracbc+a+fracca+bgeqfrac32$Prove when $abc=1$: $ fraca2+bc + fracb2+ca+fracc2+ab geq 1$Where does this trignometric substition come from?Show that $fraca+b2 geq sqrtab geq frac2aba+b$
$begingroup$
I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
asked Mar 27 at 23:43
hopefullyhopefully
315215
$endgroup$
add a comment |
$begingroup$
I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
asked Mar 27 at 23:43
hopefullyhopefully
315215
$endgroup$
add a comment |
$begingroup$
I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
asked Mar 27 at 23:43
hopefullyhopefully
315215
$endgroup$
I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me, please?
calculus inequality
calculus inequality
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
asked Mar 27 at 23:43
hopefullyhopefully
315215
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
asked Mar 27 at 23:43
hopefullyhopefully
315215
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
edited Mar 28 at 8:15
YuiTo Cheng
2,1863937
2,1863937
asked Mar 27 at 23:43
hopefullyhopefully
315215
asked Mar 27 at 23:43
hopefullyhopefully
315215
asked Mar 27 at 23:43
hopefullyhopefully
315215
315215
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
$$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.4k32143
$endgroup$
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1n(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
$frac1n(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
votes
active
oldest
votes
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.4k32143
$endgroup$
add a comment |
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.4k32143
$endgroup$
add a comment |
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.4k32143
$endgroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
which is the second inequality (modulo capitalization).
answered Mar 27 at 23:46
jgonjgon
16.4k32143
answered Mar 27 at 23:46
jgonjgon
16.4k32143
answered Mar 27 at 23:46
jgonjgon
16.4k32143
answered Mar 27 at 23:46
jgonjgon
16.4k32143
16.4k32143
add a comment |
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1n(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
$frac1n(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
$frac1n(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
$frac1n(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
$frac1n(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
$frac1n(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
$frac1n(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
$frac1n(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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