Mixing
Product of $left(1-fracxkright)$ with or without uniform $k$
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Is it possible to prove this:
$$left(1−fracxkright)^kge(1−A_1)cdot(1−A_2)cdots(1−A_k)$$
for x>0, and k≥1, where k is a whole number.
where $$A_1+A_2+...+A_k = x$$
At least one number in $A_1, A_2,...,A_k$ does not equal to the other.
calculus inequality
edited Jul 21 at 20:16
mrtaurho
700219
asked Jul 21 at 19:44
CSDUG
134
add a comment |Â
up vote
0
down vote
favorite
Is it possible to prove this:
$$left(1−fracxkright)^kge(1−A_1)cdot(1−A_2)cdots(1−A_k)$$
for x>0, and k≥1, where k is a whole number.
where $$A_1+A_2+...+A_k = x$$
At least one number in $A_1, A_2,...,A_k$ does not equal to the other.
calculus inequality
edited Jul 21 at 20:16
mrtaurho
700219
asked Jul 21 at 19:44
CSDUG
134
1
Try the AM-GM inequality.
– Mike Earnest
Jul 21 at 19:54
1
What exactly do you mean with your numbers $A1,A2,...,Ak$? Do they stand for multiplies of $A$, such as $Acdot 1, Acdot 2,...,Acdot k$, or are the numbers just indices, like $A_1,A_2,...,A_k$?
– mrtaurho
Jul 21 at 20:03
numbers are just indices
– CSDUG
Jul 21 at 20:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is it possible to prove this:
$$left(1−fracxkright)^kge(1−A_1)cdot(1−A_2)cdots(1−A_k)$$
for x>0, and k≥1, where k is a whole number.
where $$A_1+A_2+...+A_k = x$$
At least one number in $A_1, A_2,...,A_k$ does not equal to the other.
calculus inequality
edited Jul 21 at 20:16
mrtaurho
700219
asked Jul 21 at 19:44
CSDUG
134
Is it possible to prove this:
$$left(1−fracxkright)^kge(1−A_1)cdot(1−A_2)cdots(1−A_k)$$
for x>0, and k≥1, where k is a whole number.
where $$A_1+A_2+...+A_k = x$$
At least one number in $A_1, A_2,...,A_k$ does not equal to the other.
calculus inequality
edited Jul 21 at 20:16
mrtaurho
700219
asked Jul 21 at 19:44
CSDUG
134
edited Jul 21 at 20:16
mrtaurho
700219
edited Jul 21 at 20:16
mrtaurho
700219
edited Jul 21 at 20:16
mrtaurho
700219
700219
asked Jul 21 at 19:44
CSDUG
134
asked Jul 21 at 19:44
CSDUG
134
asked Jul 21 at 19:44
CSDUG
134
134
1
Try the AM-GM inequality.
– Mike Earnest
Jul 21 at 19:54
1
What exactly do you mean with your numbers $A1,A2,...,Ak$? Do they stand for multiplies of $A$, such as $Acdot 1, Acdot 2,...,Acdot k$, or are the numbers just indices, like $A_1,A_2,...,A_k$?
– mrtaurho
Jul 21 at 20:03
numbers are just indices
– CSDUG
Jul 21 at 20:08
add a comment |Â
1
Try the AM-GM inequality.
– Mike Earnest
Jul 21 at 19:54
1
What exactly do you mean with your numbers $A1,A2,...,Ak$? Do they stand for multiplies of $A$, such as $Acdot 1, Acdot 2,...,Acdot k$, or are the numbers just indices, like $A_1,A_2,...,A_k$?
– mrtaurho
Jul 21 at 20:03
numbers are just indices
– CSDUG
Jul 21 at 20:08
1
1
Try the AM-GM inequality.
– Mike Earnest
Jul 21 at 19:54
Try the AM-GM inequality.
– Mike Earnest
Jul 21 at 19:54
1
1
What exactly do you mean with your numbers $A1,A2,...,Ak$? Do they stand for multiplies of $A$, such as $Acdot 1, Acdot 2,...,Acdot k$, or are the numbers just indices, like $A_1,A_2,...,A_k$?
– mrtaurho
Jul 21 at 20:03
What exactly do you mean with your numbers $A1,A2,...,Ak$? Do they stand for multiplies of $A$, such as $Acdot 1, Acdot 2,...,Acdot k$, or are the numbers just indices, like $A_1,A_2,...,A_k$?
– mrtaurho
Jul 21 at 20:03
numbers are just indices
– CSDUG
Jul 21 at 20:08
numbers are just indices
– CSDUG
Jul 21 at 20:08
add a comment |Â
1 Answer
1
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1
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$$left(1-fracxkright)^k~geq~(1-A_1)(1-A_2)cdots(1-A_k)$$
By using the condition $A_1+A_2+ cdots + A_k~=~x$ the inequality becomes
$$beginalign
(1-A_1)(1-A_2)cdots(1-A_k)~&leq~left(1-frac(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frack-(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)kright)^k\
sqrt[hugek](1-A_1)(1-A_2)cdots(1-A_k)~&leq~frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)k
endalign$$
The L.H.S. of this inequality is just the geometric mean of the $k$-factors $(1-A_k)$ and the R.H.S. is the arithmetic mean of the $k$-summands $(1-A_k)$. The GM-AM inequalitiy is known to be true and so the given is.
answered Jul 21 at 20:37
mrtaurho
700219
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$left(1-fracxkright)^k~geq~(1-A_1)(1-A_2)cdots(1-A_k)$$
By using the condition $A_1+A_2+ cdots + A_k~=~x$ the inequality becomes
$$beginalign
(1-A_1)(1-A_2)cdots(1-A_k)~&leq~left(1-frac(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frack-(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)kright)^k\
sqrt[hugek](1-A_1)(1-A_2)cdots(1-A_k)~&leq~frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)k
endalign$$
The L.H.S. of this inequality is just the geometric mean of the $k$-factors $(1-A_k)$ and the R.H.S. is the arithmetic mean of the $k$-summands $(1-A_k)$. The GM-AM inequalitiy is known to be true and so the given is.
answered Jul 21 at 20:37
mrtaurho
700219
add a comment |Â
up vote
1
down vote
$$left(1-fracxkright)^k~geq~(1-A_1)(1-A_2)cdots(1-A_k)$$
By using the condition $A_1+A_2+ cdots + A_k~=~x$ the inequality becomes
$$beginalign
(1-A_1)(1-A_2)cdots(1-A_k)~&leq~left(1-frac(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frack-(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)kright)^k\
sqrt[hugek](1-A_1)(1-A_2)cdots(1-A_k)~&leq~frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)k
endalign$$
The L.H.S. of this inequality is just the geometric mean of the $k$-factors $(1-A_k)$ and the R.H.S. is the arithmetic mean of the $k$-summands $(1-A_k)$. The GM-AM inequalitiy is known to be true and so the given is.
answered Jul 21 at 20:37
mrtaurho
700219
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$left(1-fracxkright)^k~geq~(1-A_1)(1-A_2)cdots(1-A_k)$$
By using the condition $A_1+A_2+ cdots + A_k~=~x$ the inequality becomes
$$beginalign
(1-A_1)(1-A_2)cdots(1-A_k)~&leq~left(1-frac(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frack-(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)kright)^k\
sqrt[hugek](1-A_1)(1-A_2)cdots(1-A_k)~&leq~frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)k
endalign$$
The L.H.S. of this inequality is just the geometric mean of the $k$-factors $(1-A_k)$ and the R.H.S. is the arithmetic mean of the $k$-summands $(1-A_k)$. The GM-AM inequalitiy is known to be true and so the given is.
answered Jul 21 at 20:37
mrtaurho
700219
$$left(1-fracxkright)^k~geq~(1-A_1)(1-A_2)cdots(1-A_k)$$
By using the condition $A_1+A_2+ cdots + A_k~=~x$ the inequality becomes
$$beginalign
(1-A_1)(1-A_2)cdots(1-A_k)~&leq~left(1-frac(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frack-(A_1+A_2+ cdots + A_k)kright)^k\
&leq~left(frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)kright)^k\
sqrt[hugek](1-A_1)(1-A_2)cdots(1-A_k)~&leq~frac(1-A_1)+(1-A_2)+ cdots + (1-A_k)k
endalign$$
The L.H.S. of this inequality is just the geometric mean of the $k$-factors $(1-A_k)$ and the R.H.S. is the arithmetic mean of the $k$-summands $(1-A_k)$. The GM-AM inequalitiy is known to be true and so the given is.
answered Jul 21 at 20:37
mrtaurho
700219
answered Jul 21 at 20:37
mrtaurho
700219
answered Jul 21 at 20:37
mrtaurho
700219
answered Jul 21 at 20:37
mrtaurho
700219
700219
add a comment |Â
add a comment |Â
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1
Try the AM-GM inequality.
– Mike Earnest
Jul 21 at 19:54
1
What exactly do you mean with your numbers $A1,A2,...,Ak$? Do they stand for multiplies of $A$, such as $Acdot 1, Acdot 2,...,Acdot k$, or are the numbers just indices, like $A_1,A_2,...,A_k$?
– mrtaurho
Jul 21 at 20:03
numbers are just indices
– CSDUG
Jul 21 at 20:08