Arithmetic mean geometric mean inequality unclearproving inequality?Practicing the arithmetic-geometric means...
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Arithmetic mean geometric mean inequality unclear
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Arithmetic mean geometric mean inequality unclear
proving inequality?Practicing the arithmetic-geometric means inequalityArithmetic Mean and Geometric Mean Question, Guidance NeededHow prove Reversing the Arithmetic mean – Geometric mean inequality？Mean Value Theorem and Inequality.Using arithmetic mean>geometric meanNesbitt's Inequality $frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b}geqfrac{3}{2}$Problem in Arithmetic Mean - Geometric Mean inequalityProving Cauchy-Schwarz with Arithmetic Geometric meanInequality involving a kind of Harmonic mean
$begingroup$
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 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
$endgroup$
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 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
$endgroup$
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 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
$endgroup$
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 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
edited 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
edited 5 hours ago
Bernard
123k741117
edited 5 hours ago
Bernard
123k741117
edited 5 hours ago
Bernard
123k741117
123k741117
asked 5 hours ago
hopefullyhopefully
274114
asked 5 hours ago
hopefullyhopefully
274114
asked 5 hours ago
hopefullyhopefully
274114
274114
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 5 hours ago
jgonjgon
16k32143
$endgroup$
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 5 hours ago
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
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered 5 hours ago
jgonjgon
16k32143
$endgroup$
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 5 hours ago
jgonjgon
16k32143
$endgroup$
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 5 hours ago
jgonjgon
16k32143
$endgroup$
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 5 hours ago
jgonjgon
16k32143
answered 5 hours ago
jgonjgon
16k32143
answered 5 hours ago
jgonjgon
16k32143
answered 5 hours ago
jgonjgon
16k32143
16k32143
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 5 hours ago
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 5 hours ago
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 5 hours ago
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 5 hours ago
marty cohenmarty cohen
74.9k549130
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
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