Stronger than AM-GM and a conjectureQuestion about Rudin's Theorem 8.22 proof (Stirling's formula) - part...
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Stronger than AM-GM and a conjecture
Question about Rudin's Theorem 8.22 proof (Stirling's formula) - part (b)If $a_1a_2cdots a_n =1$, then $prod_{i=1}^n (1+a_i) geq 2^n$A conjecture similar to the Hardy inequalityA inequality about $x_1,x_2,ldots, x_n$Prove that for a family of $100n$ subsets of a $[n]$ such that each subset has size $m$, there exists two whose intersection is at least $m^2/10n$.Hard inequality with condition ($xyz=1$)Inequality using JensenRefinement of a strong inequalityPower sum inequalityA nice power sum inequalityA conjecture about power sum : $e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$ and $a+b+c=3$
$begingroup$
Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$
A possible way to generalize this is the following conjecture:
Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$
Any hints would be appreciable to solve this. Thanks in advance.
real-analysis inequality contest-math a.m.-g.m.-inequality
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
$endgroup$
add a comment |
$begingroup$
Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$
A possible way to generalize this is the following conjecture:
Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$
Any hints would be appreciable to solve this. Thanks in advance.
real-analysis inequality contest-math a.m.-g.m.-inequality
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
$endgroup$
1
$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47
6
$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54
$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48
$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13
$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25
add a comment |
$begingroup$
Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$
A possible way to generalize this is the following conjecture:
Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$
Any hints would be appreciable to solve this. Thanks in advance.
real-analysis inequality contest-math a.m.-g.m.-inequality
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
$endgroup$
Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$
A possible way to generalize this is the following conjecture:
Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$
Any hints would be appreciable to solve this. Thanks in advance.
real-analysis inequality contest-math a.m.-g.m.-inequality
real-analysis inequality contest-math a.m.-g.m.-inequality
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
edited Feb 28 at 10:17
GNUSupporter 8964民主女神 地下教會
14k82650
14k82650
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
asked Feb 28 at 7:37
FatsWallersFatsWallers
1307
1307
1
$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47
6
$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54
$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48
$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13
$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25
add a comment |
1
$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47
6
$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54
$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48
$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13
$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25
1
1
$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47
$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47
6
6
$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54
$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54
$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48
$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48
$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13
$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13
$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25
$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$
The generalization is false, if my calculations are correct. Take
begin{align*}
n&=4\
a_1&=tfrac12\
a_i&=1qquad(2leq ileq 4)\
f(x)&=begin{cases}x/e^2& (xleq 7/8)\
7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
end{cases}\
f(a_1)&=1/2e^2\
f(a_i)&=1/eqquad(2leq ileq 4)\
A&=tfrac1nsum f(a_i)approx0.29283\
G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
end{align*}
Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$
answered Mar 15 at 10:16
DapDap
19k842
$endgroup$
add a comment |
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$begingroup$
For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$
The generalization is false, if my calculations are correct. Take
begin{align*}
n&=4\
a_1&=tfrac12\
a_i&=1qquad(2leq ileq 4)\
f(x)&=begin{cases}x/e^2& (xleq 7/8)\
7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
end{cases}\
f(a_1)&=1/2e^2\
f(a_i)&=1/eqquad(2leq ileq 4)\
A&=tfrac1nsum f(a_i)approx0.29283\
G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
end{align*}
Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$
answered Mar 15 at 10:16
DapDap
19k842
$endgroup$
add a comment |
$begingroup$
For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$
The generalization is false, if my calculations are correct. Take
begin{align*}
n&=4\
a_1&=tfrac12\
a_i&=1qquad(2leq ileq 4)\
f(x)&=begin{cases}x/e^2& (xleq 7/8)\
7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
end{cases}\
f(a_1)&=1/2e^2\
f(a_i)&=1/eqquad(2leq ileq 4)\
A&=tfrac1nsum f(a_i)approx0.29283\
G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
end{align*}
Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$
answered Mar 15 at 10:16
DapDap
19k842
$endgroup$
add a comment |
$begingroup$
For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$
The generalization is false, if my calculations are correct. Take
begin{align*}
n&=4\
a_1&=tfrac12\
a_i&=1qquad(2leq ileq 4)\
f(x)&=begin{cases}x/e^2& (xleq 7/8)\
7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
end{cases}\
f(a_1)&=1/2e^2\
f(a_i)&=1/eqquad(2leq ileq 4)\
A&=tfrac1nsum f(a_i)approx0.29283\
G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
end{align*}
Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$
answered Mar 15 at 10:16
DapDap
19k842
$endgroup$
For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$
The generalization is false, if my calculations are correct. Take
begin{align*}
n&=4\
a_1&=tfrac12\
a_i&=1qquad(2leq ileq 4)\
f(x)&=begin{cases}x/e^2& (xleq 7/8)\
7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
end{cases}\
f(a_1)&=1/2e^2\
f(a_i)&=1/eqquad(2leq ileq 4)\
A&=tfrac1nsum f(a_i)approx0.29283\
G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
end{align*}
Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$
answered Mar 15 at 10:16
DapDap
19k842
edited Mar 15 at 10:27
answered Mar 15 at 10:16
DapDap
19k842
answered Mar 15 at 10:16
DapDap
19k842
answered Mar 15 at 10:16
DapDap
19k842
19k842
add a comment |
add a comment |
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1
$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47
6
$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54
$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
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– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48
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@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
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– Martin R
Feb 28 at 10:13
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@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
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– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25